( ) [2] H 4 4! H 4 4! (5 4 3 )= = Fortran C 0 #include <stdio.h> 1 #include
|
|
|
- ときな うばら
- 5 years ago
- Views:
Transcription
1 J.JSSAC (2006) Vol. 12, No. 3, pp IIJ bit TULIPS KING KISS WINK 1 ( ) Ackermann GC [1] CPU 6502 wada@u-tokyo.ac.jp c 2006 Japan Society for Symbolic and Algebraic Computation
2 ( ) [2] H 4 4! H 4 4! (5 4 3 )= = Fortran C 0 #include <stdio.h> 1 #include <math.h> 2 main (){ 3 float e[93]; 4 int ir[60], ib[3][6], ia[4][24], kc[24]; 5 float test=10.0,a,b,c,d,apb,amb,atb,cpd,cmd,ctd,bs,cs,ds, 6 cpds,cmds,asb,bscpd,bscmd,ss,bpcs,bmcs; 7 int kk,ll,mm,nn,i,j,k,l,m,n,ka,kb; 8 printf("\n"); 9 for(kk=1;kk<=10;kk++) 10 for(ll=kk;ll<=10;ll++) 11 for(mm=ll;mm<=10;mm++) 12 for(nn=mm;nn<=10;nn++){ 13 k=kk-1;l=ll-1;m=mm-1;n=nn-1; 14 printf("********%3d%3d%3d%3d ********\n",k,l,m,n); 15 for(j=0;j<3;j++){ib[j][j]=l;ib[j][j+3]=l;} 16 ib[1][0]=m;ib[0][1]=m;ib[0][2]=m;ib[2][3]=m;ib[2][4]=m;ib[1][5]=m; 17 ib[2][0]=n;ib[2][1]=n;ib[1][2]=n;ib[1][3]=n;ib[0][4]=n;ib[0][5]=n; 18 n=0; 19 for(l=0;l<6;l++) 20 for(ka=0;ka<4;ka++){ 21 kb=0; 22 for(m=0;m<4;m++) 23 if(m==ka)ia[m][n]=k;else{ia[m][n]=ib[kb][l];kb=kb+1;} 24 n=n+1;}
3 J.JSSAC Vol. 12, No. 3, /* permutation of a,b,c,d*/ 26 for(l=0;l<24;l++){ 27 kc[l]=0; 28 for(m=0;m<4;m++)kc[l]=kc[l]*10+ia[m][l];} 29 for(l=0;l<23;l++){ 30 m=l+1; 31 j=kc[l]; 32 if(l>=0)for(n=m;n<24;n++)if(kc[n]==j)kc[n]=-1;} 33 /* start testing */ 34 for(kb=0;kb<24;kb++)if(kc[kb]>=0){ 35 a=ia[0][kb]; b=ia[1][kb]; c=ia[2][kb]; d=ia[3][kb]; 36 apb=a+b; amb=a-b; atb=a*b; cpd=c+d; cmd=c-d; ctd=c*d; 37 bs=b; cs=c; ds=d; 38 if(b==0.0)bs= ; 39 if(c==0.0)cs= ; 40 if(d==0.0)ds= ; 41 cpds=cpd;cmds=cmd;asb=a/bs;bscpd=b/cs+d;bscmd=b/cs-d;ss= ; 42 if(cpd==0.0)cpds=ss; 43 if(cmd==0.0)cmds=ss; 44 if(bscpd==0.0)bscpd=ss; 45 if(bscmd==0.0)bscmd=ss; 46 bpcs=b+c;bmcs=b-c; 47 if(bpcs==0.0)bpcs=ss; 48 if(bmcs==0.0)bmcs=ss; 49 /* evaluate expressions */ 50 e[0] = apb + cpd; e[1] = apb + cmd; 51 e[2] = (apb + c) * d; e[3] = (apb + c) / ds; 52 e[4] = apb - cpd; e[5] = (apb - c) * d; 53 e[6] = (apb - c) / ds; e[7] = apb * c + d; 54 e[8] = apb * c - d; e[9] = apb * ctd; 55 e[10] = apb * c / ds; e[11] = apb / cs + d; 56 e[12] = apb / cs - d; e[13] = apb / cs / ds; 57 e[14] = (amb - c) * d; e[15] = amb * c + d; 58 e[16] = amb * c - d; e[17] = amb * c * d; 59 e[18] = amb * c / ds; e[19] = amb / cs + d; 60 e[20] = atb + cpd; e[21] = atb + cmd; 61 e[22] = (atb + c) * d; e[23] = (atb + c) / ds;
4 e[24] = atb - cpd; e[25] = (atb - c) * d; 63 e[26] = (atb - c) / ds; e[27] = atb * c + d; 64 e[28] = atb * c - d; e[29] = atb * ctd; 65 e[30] = atb * c / ds; e[31] = atb / cs + d; 66 e[32] = atb / cs - d; e[33] = atb / cs / ds; 67 e[34] = asb + cpd; e[35] = asb + cmd; 68 e[36] = (asb + c) * d; e[37] = (asb + c) / ds; 69 e[38] = (asb - c) * d; e[39] = asb / cs + d; 70 e[40] = apb - ctd; e[41] = apb - c / ds; 71 e[42] = apb * cpd; e[43] = apb * cmd; 72 e[44] = apb / cpds; e[45] = apb / cmds; 73 e[46] = amb * cmd; e[47] = atb + ctd; 74 e[48] = atb + c / ds; e[49] = atb - ctd; 75 e[50] = atb - c / ds; e[51] = atb / cpds; 76 e[52] = atb / cmds; e[53] = asb + c / ds; 77 e[54] = a / bscpd; e[55] = a / bscmd; 78 e[56] = (a - b * c) * d; e[57] = (a - b / cs) * d; 79 e[58] = a / bpcs + d; e[59] = a / bmcs + d; 80 e[60] = - a / bscmd; 81 /* start testing */ 82 k=0; 83 for(i=0;i<61;i++)if(fabs(e[i]-test)<0.0001){ir[k]=i;k=k+1;} 84 if(k>0){ 85 for(n=0;n<4;n++)printf("%2d ",ia[n][kb]); 86 for(j=0;j<k;j++)printf("%3d ",ir[j]);printf("\n");}}}} IBM 7040 WATFOR test= kk ll mm nn kk Fortran 1 1 k l m n (13 ) 15 4 permutation [3] LISP [4] M Lisp
5 J.JSSAC Vol. 12, No. 3, comb1[a;x;y]= [null[y]->cons[append[a; cons[x;y]];nil]; t->cons[append[a;cons[car[y];nil]]; comb1[append[a;cons[car[y];nil]]; x;cdr[y]]]]]; comb2[x;y]=[null[y]->nil; t->append[comb1[nil;x;car[y]]; comb2[x;cdr[y]]]]; perm[x]=[null[cdr[x]]->cons[x;nil]; t->comb2[car[x];perm[cdr x]]]]. LISP (a b c) (b c) ((b c) (c b)) (comb2 ) a (comb1 ) Lisp 4 Fortran l m n 3 (15 17 ) ib 2 ((m n) (n m)) l ib l m m l n n 1 m l n n l m 2 n n l m m l k 4 ia (18 24 ) ia k l l l k m m m k m n 1 l k m m m k l l m k m 2 m m k n l l k n n n l 3 n n n k n n n k l l k Knuth TAOCP 4 2 Plain Change
6 k l m n 0, 0, 0, 0 kc kc kc 1 (32 ) 1 34 kc >= 0 (0, 0, 0, 0 ) ia kb a b c d (35 ) e e 0 (+ (+ a b) (+ c d)) 1 (+ (+ a b) (- c d)) 2 (* (+ (+ a b) c) d) 3 (/ (+ (+ a b) c) d) 4 (- (+ a b) (+ c d)) 5 (* (- (+ a b) c) d) 6 (/ (- (+ a b) c) d) 7 (+ (* (+ a b) c) d) 8 (- (* (+ a b) c) d) 9 (* (+ a b) (* c d)) 10 (/ (* (+ a b) c) d) 11 (+ (/ (+ a b) c) d) 12 (- (/ (+ a b) c) d) 13 (/ (/ (+ a b) c) d) 14 (* (- (- a b) c) d) 15 (+ (* (- a b) c) d) 16 (- (* (- a b) c) d) 17 (* (* (- a b) c) d) 18 (/ (* (- a b) c) d) 19 (+ (/ (- a b) c) d) 20 (+ (* a b) (+ c d)) 21 (+ (* a b) (- c d)) 22 (* (+ (* a b) c) d) 23 (/ (+ (* a b) c) d) 24 (- (* a b) (+ c d)) 25 (* (- (* a b) c) d) 26 (/ (- (* a b) c) d) 27 (+ (* (* a b) c) d) 28 (- (* (* a b) c) d) 29 (* (* a b) (* c d)) 30 (/ (* (* a b) c) d) 31 (+ (/ (* a b) c) d) 32 (- (/ (* a b) c) d) 33 (/ (/ (* a b) c) d) 34 (+ (/ a b) (+ c d)) 35 (+ (/ a b) (- c d)) 36 (* (+ (/ a b) c) d) 37 (/ (+ (/ a b) c) d) 38 (* (- (/ a b) c) d) 39 (+ (/ (/ a b) c) d) 40 (- (+ a b) (* c d)) 41 (- (+ a b) (/ c d)) 42 (* (+ a b) (+ c d)) 43 (* (+ a b) (- c d)) 44 (/ (+ a b) (+ c d)) 45 (/ (+ a b) (- c d)) 46 (* (- a b) (- c d)) 47 (+ (* a b) (* c d)) 48 (+ (* a b) (/ c d)) 49 (- (* a b) (* c d)) 50 (- (* a b) (/ c d)) 51 (/ (* a b) (+ c d)) 52 (/ (* a b) (- c d)) 53 (+ (/ a b) (/ c d)) 54 (/ a (+ (/ b c) d)) 55 (/ a (- (/ b c) d)) 56 (* (- a (* b c)) d) 57 (* (- a (/ b c)) d) 58 (+ (/ a (+ b c)) d) 59 (+ (/ a (- b c)) d) 60 (- (/ a (- (/ b c) d))) 2 4
7 J.JSSAC Vol. 12, No. 3, (0. 18) (+ a b) (0. 18) (* a b) (0. 81) (- a b) ( 9. 9) (/ a b) (0. 9) (+ (+ a b) c)) (0. 27) (* (* a b) c)) (0. 729) (+ (* a b) c)) (0. 90) (* (+ a b) c)) (0. 162) (+ (/ a b) c)) (0. 18) (* (- a b) c)) ( ) (- (+ a b) c)) ( 9. 18) (/ (* a b) c)) (0. 81) (- (- a b) c)) ( 18. 9) (/ (/ a b) c)) (0. 9) (- (* a b) c)) ( 9. 81) (/ (+ a b) c)) (0. 18) (- (/ a b) c)) ( 9. 9) (/ (- a b) c)) ( 9. 9) (- a (* b c))) ( 81. 9) (/ a (+ b c))) (0. 9) (- a (/ b c))) ( 9. 9) (/ a (- b c))) ( 9. 9) 0 (+ (+ (+ a b) c) d) (0. 36) 29 (* (* (* a b) c) d) ( ) 20 (+ (+ (* a b) c) d) (0. 99) 9 (* (* (+ a b) c) d) ( ) 34 (+ (+ (/ a b) c) d) (0. 27) 17 (* (* (- a b) c) d) ( ) 27 (+ (* (* a b) c) d) (0. 738) 2 (* (+ (+ a b) c) d) (0. 243) 7 (+ (* (+ a b) c) d) (0. 171) 22 (* (+ (* a b) c) d) (0. 810) 15 (+ (* (- a b) c) d) ( ) 36 (* (+ (/ a b) c) d) (0. 162) 31 (+ (/ (* a b) c) d) (0. 90) 5 (* (- (+ a b) c) d) ( ) 39 (+ (/ (/ a b) c) d) (0. 18) 14 (* (- (- a b) c) d) ( ) 11 (+ (/ (+ a b) c) d) (0. 27) 25 (* (- (* a b) c) d) ( ) 19 (+ (/ (- a b) c) d) ( 9. 18) 38 (* (- (/ a b) c) d) ( ) 58 (+ (/ a (+ b c)) d) (0. 18) 56 (* (- a (* b c)) d) ( ) 59 (+ (/ a (- b c)) d) ( 9. 18) 57 (* (- a (/ b c)) d) ( ) 47 (+ (* a b) (* c d)) (0. 162) 42 (* (+ a b) (+ c d)) (0. 324) 48 (+ (* a b) (/ c d)) (0. 90) 43 (* (+ a b) (- c d)) ( ) 53 (+ (/ a b) (/ c d)) (0. 18) 46 (* (- a b) (- c d)) ( ) 1 (- (+ (+ a b) c) d) ( 9. 27) 30 (/ (* (* a b) c) d) (0. 729) 21 (- (+ (* a b) c) d) ( 9. 90) 10 (/ (* (+ a b) c) d) (0. 162) 35 (- (+ (/ a b) c) d) ( 9. 18) 18 (/ (* (- a b) c) d) ( )
8 (- (- (+ a b) c) d) ( ) 33 (/ (/ (* a b) c) d) (0. 81) (- (- (- a b) c) d) ( 27. 9) (/ (/ (/ a b) c) d) (0. 9) 24 (- (- (* a b) c) d) ( ) 13 (/ (/ (+ a b) c) d) (0. 18) (- (- (/ a b) c) d) ( 18. 9) (/ (/ (- a b) c) d) ( 9. 9) (- (- a (* b c)) d) ( 90. 9) (/ (/ a (+ b c)) d) (0. 9) (- (- a (/ b c)) d) ( 18. 9) (/ (/ a (- b c)) d) ( 9. 9) 28 (- (* (* a b) c) d) ( ) 3 (/ (+ (+ a b) c) d) (0. 27) 8 (- (* (+ a b) c) d) ( ) 23 (/ (+ (* a b) c) d) (0. 90) 16 (- (* (- a b) c) d) ( ) 37 (/ (+ (/ a b) c) d) (0. 18) 32 (- (/ (* a b) c) d) ( 9. 81) 6 (/ (- (+ a b) c) d) ( 9. 18) (- (/ (/ a b) c) d) ( 9. 9) (/ (- (- a b) c) d) ( 18. 9) 12 (- (/ (+ a b) c) d) ( 9. 18) 26 (/ (- (* a b) c) d) ( 9. 81) (- (/ (- a b) c) d) ( 18. 9) (/ (- (/ a b) c) d) ( 9. 9) (- (/ a (+ b c)) d) ( 9. 9) (/ (- a (* b c)) d) ( 81. 9) (- (/ a (- b c)) d) ( 18. 9) (/ (- a (/ b c)) d) ( 9. 9) (- a (* (* b c) d)) ( ) (/ a (+ (+ b c) d)) (0. 9) (- a (* (+ b c) d)) ( ) (/ a (+ (* b c) d)) (0. 9) (15)(- a (* (- b c) d)) ( ) 54 (/ a (+ (/ b c) d)) (0. 81) (- a (/ (* b c) d)) ( 81. 9) (/ a (- (+ b c) d)) ( 9. 9) (- a (/ (/ b c) d)) ( 9. 9) (/ a (- (- b c) d)) ( 9. 9) (- a (/ (+ b c) d)) ( 18. 9) (/ a (- (* b c) d)) ( 9. 9) (19)(- a (/ (- b c) d)) ( 9. 18) 55 (/ a (- (/ b c) d)) ( ) (- a (/ b (+ c d))) ( 9. 9) (/ a (- b (* c d))) ( 9. 9) (59)(- a (/ b (- c d))) ( 9. 18) 60 (/ a (- b (/ c d))) ( ) 40 (- (+ a b) (* c d)) ( ) 51 (/ (* a b) (+ c d)) (0. 81) 41 (- (+ a b) (/ c d)) ( 9. 18) 52 (/ (* a b) (- c d)) ( ) 49 (- (* a b) (* c d)) ( ) 44 (/ (+ a b) (+ c d)) (0. 18) 50 (- (* a b) (/ c d)) ( 9. 81) 45 (/ (+ a b) (- c d)) ( ) (- (/ a b) (* c d)) ( 81. 9) (/ (- a b) (+ c d)) ( 9. 9) (- (/ a b) (/ c d)) ( 9. 9) (/ (- a b) (- c d)) ( 9. 9) 2 + * - / a + b a - b apb amb
9 J.JSSAC Vol. 12, No. 3, b c d bs cs ds asb( adb a slash b ) 60 (/ a (- b (/ c d))) (- (/ a (- (/ b c) d))) (/ 8 (- 1 (/ 1 5))) 10 a (- (/ b c) d) bscmd 50 e e[0] a + b + c + d e[60] e ( 10 ) TEST (setq ops (+ - * :)) (mapcar ops (lambda (x) (mapcar ops (lambda (y) (mapcar ops (lambda (z) (princ (,x (terpri) (princ (,x (terpri) (princ @))) (terpri) (princ @))) (terpri) (princ @)))) (+ (+ (+ (+ utilisp match back-
10 slash(\) colon(:) match clause (defun mod3 (x) ;3 (lets ((y (match x (( + ( - a b) c) (- (+,a,c),b)) (( (b c d)) (+ (( - a ( + b c)) (- (-,a,b),c)) (( - a ( - b c)) (+ (-,a,b),c)) ; (t x)))) (cond ((equal x y) y) (t (mod3 y))))) (defun mod22 (x) @)) (match x (( + a) (( + a) (( + a (( + a (( + @))) (( @)) @))) (( - a) (( - a (( - a ; (t x))) (defun modsub (x) (cond ((null x) x) ((atom (cadr x)) (list (car x) (cadr x) (mod3 (caddr x)))) ((atom (caddr x)) (list (car x) (mod3 (cadr x)) (caddr x))) (t (mod22 x)))) (defun mod (foo) (setq exs ()) (mapcar foo (lambda (ex0) (princ ex0) (lets ((ex1 (mod22 ex0))
11 J.JSSAC Vol. 12, No. 3, (ex2 (modsub ex1)) (ex3 (match ex2 (( a) (( a) (t ex2))) (ex4 (mod3 ex3)) (ex5 (modsub ex4))) (princ ex5) (terpri) (setq z (assoc ex5 exs)) (cond (z nil) (t (setq exs (cons (list ex5) exs))))))) (mapcar (reverse exs) (lambda (x) (princ x) (terpri)))) (10 93 ) Ackermann Ackermann (define (A x y) (cond ((= x 0) (+ y 1)) ((= y 0) (A (- x 1) 1)) (else (A (- x 1) (A x (- y 1)))))) Ackermann 0 x 10, 0 y Fortran [5] INTEGER FUNCTION F(IX,IY) DIMENSION M(21) DO 1 I=1,21 1 M(I)=1-2*MOD(I,2) 2 F=M(1)+2 DO 3 I=1,21,2 M(I)=M(I)+1 Sussman Ackermann
12 IF(I.GT.2*IX.OR.M(I).LT.M(I+1))IF(M(2*IX+1)-IY)2,4,4 3 M(I+1)=F 4 RETURN END M ys bs F bs[0] bs[i]=a(ys[i],i) ys bs y C 6 Ackermann 0 nack(int x,int y){ F M1 M2 M3 M4 M5 1 int i,ys[11],bs[11]; i bs0 ys0 bs1 ys1 bs2 ys2 2 for(i=0;i<11;i++)ys[i]=-1; for(i=1;i<11;i++)bs[i]=1; l1: bs[0]=ys[0]+2;i=0; l2: ys[i]=ys[i]+1; if((i>=x) (ys[i]<bs[i+1])) {if(ys[x]>=y)goto l3; else goto l1;} bs[i+1]=bs[0];i=i+1;if(i<=10)goto l2; l3:return(bs[0]);} \x y bs0 bs1 bs2 bs A(2, 3) 9 A(2, 2) 7, A(1, 7) 9 A(1, 6) 8 A(0, 8) A(2, 2) 7 A(2, 1)
13 J.JSSAC Vol. 12, No. 3, bs2 7 ys2 2 A(2, 2) = 7 ys1 7 bs ys (C 5 ) ys1 7 bs1(=9) bs2 C (8 )bs[0] bs bs1=8 ys1=6 18 bs0=9 ys0=8 bs[i] ys[i] 1 bs[i+1] ys[i] ys[i]<bs[i+1] bs[0] (4 ) bs[i+1] i l2 ys 1 Ackermann y = 0 1 y = 1 bs 1 1 ys 0 ys 1 bs ys Ackermann Facom K KLISP Tosbac KT-Pilot K (KLISP 16K ) ( )Fortran Lisp IBM ( ) FORTRAN Knuth TAOCP Knuth LISP M- KLISP M-
14 [1] : GC, ( ), [2] :,,Vol.11,No.7,pp ( ). [3], :,, Vol.10,No.2( ). [4] : LISP [5] System-5:,, Vol.11,No.4,pp.65-67( ). [6] : KLISP,, Vol.11,No.10, pp
r3.dvi
2012 3 / Lisp(2) 2012.4.19 1 Lisp 1.1 Lisp Lisp (1) (setq) (2) (3) setq defun (defun (... &aux...)...) ( ) ( nil ) [1]> (defun sisoku (x y &aux wa sa sho seki) (setq wa (+ x y)) (setq sa (- x y)) (setq
r3.dvi
/ 94 2 (Lisp ) 3 ( ) 1994.5.16,1994.6.15 1 cons cons 2 >(cons a b) (A. B).? Lisp (S ) cons 2 car cdr n A B C D nil = (A B C D) nil nil A D E = (A (B C) D E) B C E = (A B C D. E) A B C D B = (A. B) A nil.
1. A0 A B A0 A : A1,...,A5 B : B1,...,B
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A B f : A B 4 (i) f (ii) f (iii) C 2 g, h: C A f g = f h g = h (iv) C 2 g, h: B C g f = h f g = h 4 (1) (i) (iii) (2) (iii) (i) (3) (ii) (iv) (4)
Microsoft PowerPoint - 2-LispProgramming-full
2013 年 5 月 31 日 Lisp プログラミング入門 西田豊明 Copyright 2013 Toyoaki Nishida All Rights Reserved. 今回あらすじ 1. Lisp の実践的な使い方を学習する. 2. Lisp インタープリタの動かし方, 電卓的使い方, 関数定義, 条件分岐,S 式の基本操作, プログラミング手法, プロトタイピング法などを中心に解説する.
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 5 3. 4. 5. A0 (1) A, B A B f K K A ϕ 1, ϕ 2 f ϕ 1 = f ϕ 2 ϕ 1 = ϕ 2 (2) N A 1, A 2, A 3,... N A n X N n X N, A n N n=1 1 A1 d (d 2) A (, k A k = O), A O. f
6-1
6-1 (data type) 6-2 6-3 ML, Haskell, Scala Lisp, Prolog (setq x 123) (+ x 456) (setq x "abc") (+ x 456) ; 6-4 ( ) subtype INDEX is INTEGER range -10..10; type DAY is (MON, TUE, WED, THU, FRI, SAT, SUN);
Microsoft PowerPoint Lisp.ppt [互換モード]
![Microsoft PowerPoint Lisp.ppt [互換モード]](/thumbs/95/122814430.jpg "Microsoft PowerPoint Lisp.ppt [互換モード]")
櫻井彰人 プログラム言語論 (8) Lisp, 1960 Historical Lisp 展望 古いアイデアには古いものもある 古いアイデアには新しいものもある エレガントな 極めてコンパクトな言語 C とは異世界 : 別の考え方をするよい機会 言語設計における多くの一般的課題を含む 参考文献 McCarthy, Recursive functions of symbolic expressions
1. A0 A B A0 A : A1,...,A5 B : B1,...,B
1. A0 A B A0 A : A1,...,A5 B : B1,...,B12 2. 3. 4. 5. A0 A, B Z Z m, n Z m n m, n A m, n B m=n (1) A, B (2) A B = A B = Z/ π : Z Z/ (3) A B Z/ (4) Z/ A, B (5) f : Z Z f(n) = n f = g π g : Z/ Z A, B (6)
11042 計算機言語7回目 サポートページ:
11042 7 :https://goo.gl/678wgm November 27, 2017 10/2 1(print, ) 10/16 2(2, ) 10/23 (3 ) 10/31( ),11/6 (4 ) 11/13,, 1 (5 6 ) 11/20,, 2 (5 6 ) 11/27 (7 12/4 (9 ) 12/11 1 (10 ) 12/18 2 (10 ) 12/25 3 (11
2
1 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234 12123456789012345678901234
/* do-while */ #include <stdio.h> #include <math.h> int main(void) double val1, val2, arith_mean, geo_mean; printf( \n ); do printf( ); scanf( %lf, &v
1 http://www7.bpe.es.osaka-u.ac.jp/~kota/classes/jse.html kota@fbs.osaka-u.ac.jp /* do-while */ #include #include int main(void) double val1, val2, arith_mean, geo_mean; printf( \n );
Ｃ言語によるアルゴリズムとデータ構造
Algorithms and Data Structures in C 4 algorithm List - /* */ #include List - int main(void) { int a, b, c; int max; /* */ Ÿ 3Ÿ 2Ÿ 3 printf(""); printf(""); printf(""); scanf("%d", &a); scanf("%d",
(Basic Theory of Information Processing) Fortran Fortan Fortan Fortan 1
(Basic Theory of Information Processing) Fortran Fortan Fortan Fortan 1 17 Fortran Formular Tranlator Lapack Fortran FORTRAN, FORTRAN66, FORTRAN77, FORTRAN90, FORTRAN95 17.1 A Z ( ) 0 9, _, =, +, -, *,
I ASCII ( ) NUL 16 DLE SP P p 1 SOH 17 DC1! 1 A Q a q STX 2 18 DC2 " 2 B R b
I 4 003 4 30 1 ASCII ( ) 0 17 0 NUL 16 DLE SP 0 @ P 3 48 64 80 96 11 p 1 SOH 17 DC1! 1 A Q a 33 49 65 81 97 113 q STX 18 DC " B R b 34 50 66 8 98 114 r 3 ETX 19 DC3 # 3 C S c 35 51 67 83 99 115 s 4 EOT
I. Backus-Naur BNF : N N 0 N N N N N N 0, 1 BNF N N 0 11 (parse tree) 11 (1) (2) (3) (4) II. 0(0 101)* (
2016 2016 07 28 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF : 11011 N N 0 N N 11 1001 N N N N 0, 1 BNF N N 0 11 (parse tree) 11 (1) 1100100 (2) 1111011 (3) 1110010 (4) 1001011
ex01.dvi
,. 0. 0.0. C () /******************************* * $Id: ex_0_0.c,v.2 2006-04-0 3:37:00+09 naito Exp $ * * 0. 0.0 *******************************/ #include int main(int argc, char **argv) double
203 x, y, z (x, y, z) x 6 + y 6 + z 6 = 3xyz ( 203 5) a 0, b 0, c 0 a3 + b 3 + c 3 abc 3 a = b = c 3xyz = x 6 + y 6 + z 6 = (x 2 ) 3 + (y 2 ) 3
203 24 203 x, y, z (x, y, z) x 6 + y 6 + z 6 = 3xyz ( 203 5) 202 20 a 0, b 0, c 0 a3 + b 3 + c 3 abc 3 a = b = c 3xyz = x 6 + y 6 + z 6 = (x 2 ) 3 + (y 2 ) 3 + (z 2 ) 3 3x 2 y 2 z 2 ( ) 3xyz 3(xyz) 2.
J.JSSAC Vol. 7, No. 2, Mathematica Maple,., Open asir Open xxx asir. Open xxx Open asir, asir., Open xxx, Linux Open asir Open sm1 (kan/sm1). C
J.JSSAC (1999) Vol. 7, No. 2, pp. 2-17 Open asir HPC (Received 1997/12/1) 1 Open asir Open xxx,., ( ),,,,,.,., (1) (2) (3) (4),. Open xxx,.,., 1.,.,., 0 10, dx,.,., ohara@math.kobe-u.ac.jp taka@math.kobe-u.ac.jp
XMPによる並列化実装2
2 3 C Fortran Exercise 1 Exercise 2 Serial init.c init.f90 XMP xmp_init.c xmp_init.f90 Serial laplace.c laplace.f90 XMP xmp_laplace.c xmp_laplace.f90 #include int a[10]; program init integer
[1] #include<stdio.h> main() { printf("hello, world."); return 0; } (G1) int long int float ± ±
![[1] #include<stdio.h> main() { printf(hello, world.); return 0; } (G1) int long int float ± ±](/thumbs/99/143405735.jpg "[1] #include<stdio.h> main() { printf(hello, world.); return 0; } (G1) int long int float ± ±")
[1] #include printf("hello, world."); (G1) int -32768 32767 long int -2147483648 2147483647 float ±3.4 10 38 ±3.4 10 38 double ±1.7 10 308 ±1.7 10 308 char [2] #include int a, b, c, d,
:30 12:00 I. I VI II. III. IV. a d V. VI
2017 2017 08 03 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF X [ S ] a S S ; X X X, S [, a, ], ; BNF X (parse tree) (1) [a;a] (2) [[a]] (3) [a;[a]] (4) [[a];a] : [a] X 2 222222
I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) + x * x + x x (4) * *
2015 2015 07 30 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF S + S S * S S x S +, *, x BNF S (parse tree) : * x + x x S * S x + S S S x x (1) * x x * x (2) * + x x x (3) +
:30 12:00 I. I VI II. III. IV. a d V. VI
2018 2018 08 02 10:30 12:00 I. I VI II. III. IV. a d V. VI. 80 100 60 1 I. Backus-Naur BNF N N y N x N xy yx : yxxyxy N N x, y N (parse tree) (1) yxyyx (2) xyxyxy (3) yxxyxyy (4) yxxxyxxy N y N x N yx
1 return main() { main main C 1 戻り値の型 関数名 引数 関数ブロックをあらわす中括弧 main() 関数の定義 int main(void){ printf("hello World!!\n"); return 0; 戻り値 1: main() 2.2 C main
C 2007 5 29 C 1 11 2 2.1 main() 1 FORTRAN C main() main main() main() 1 return 1 1 return main() { main main C 1 戻り値の型 関数名 引数 関数ブロックをあらわす中括弧 main() 関数の定義 int main(void){ printf("hello World!!\n"); return
Jacques Garrigue
Jacques Garrigue Garrigue 1 Garrigue 2 $ print_lines () > for i in $1; do > echo $i > done $ print_lines "a b c" a b c Garrigue 3 Emacs Lisp (defun print-lines (lines) (dolist (str lines) (insert str)
導入基礎演習.ppt
Multi-paradigm Programming Functional Programming Scheme Haskell ML Scala X10 KL1 Prolog Declarative Lang. C Procedural Lang. Java C++ Python Object-oriented Programming / (root) bin home lib 08 09
平成9年度水道事業年報 １概況 ２施設
() (mm) 12 3 31 12 3 31 4 5 6 7 8 9 10 11 12 1 2 3 145,085 146,117 146,352 146,409 146,605 146,685 146,807 147,014 147,002 147,277
08 p Boltzmann I P ( ) principle of equal probability P ( ) g ( )g ( 0 ) (4 89) (4 88) eq II 0 g ( 0 ) 0 eq Taylor eq (4 90) g P ( ) g ( ) g ( 0
08 p. 8 4 k B log g() S() k B : Boltzmann T T S k B g g heat bath, thermal reservoir... 4. I II II System I System II II I I 0 + 0 const. (4 85) g( 0 ) g ( )g ( ) g ( )g ( 0 ) (4 86) g ( )g ( 0 ) 0 (4
untitled
II yacc 005 : 1, 1 1 1 %{ int lineno=0; 3 int wordno=0; 4 int charno=0; 5 6 %} 7 8 %% 9 [ \t]+ { charno+=strlen(yytext); } 10 "\n" { lineno++; charno++; } 11 [^ \t\n]+ { wordno++; charno+=strlen(yytext);}
つくって学ぶプログラミング言語 RubyによるScheme処理系の実装
Ruby Scheme 2013-04-16 ( )! SICP *1 ;-) SchemeR SICP MIT * 1 Structure and Interpretaion of Computer Programs 2nd ed.: 2 i SchemeR Ruby Ruby Ruby Ruby & 3.0 Ruby ii https://github.com/ichusrlocalbin/scheme_in_ruby
r08.dvi
19 8 ( ) 019.4.0 1 1.1 (linked list) ( ) next ( 1) (head) (tail) ( ) top head tail head data next 1: NULL nil ( ) NULL ( NULL ) ( 1 ) (double linked list ) ( ) 1 next 1 prev 1 head cur tail head cur prev
r07.dvi
19 7 ( ) 2019.4.20 1 1.1 (data structure ( (dynamic data structure 1 malloc C free C (garbage collection GC C GC(conservative GC 2 1.2 data next p 3 5 7 9 p 3 5 7 9 p 3 5 7 9 1 1: (single linked list 1
ohp07.dvi
19 7 ( ) 2019.4.20 1 (data structure) ( ) (dynamic data structure) 1 malloc C free 1 (static data structure) 2 (2) C (garbage collection GC) C GC(conservative GC) 2 2 conservative GC 3 data next p 3 5
ex01.dvi
,. 0. 0.0. C () /******************************* * $Id: ex_0_0.c,v.2 2006-04-0 3:37:00+09 naito Exp $ * * 0. 0.0 *******************************/ #include int main(int argc, char **argv) { double
&A : A = k j 1: 4-way., A set x, way y, way y LRU y, way., A (x,y).,,, L1( 1) L2, L3 3. L1., L2,L3., TLB(Translation Lookaside Buffer). OS,. TLB, ( ),
1?,. 1,.,,. n-way (n ). 1, 4-way, n-way n (way).,., 1., ( set x ) (x), n., 2, 2 s, 2 l (, s, l )., s + l s., s,., n s. n. s + l way, (set,way)., way,. way, LRU(Least Recently Used, ). way. way, (,...).
Vol.9No.22012
6 21 2 21 8 7 118 Vol.9No.22012 6 2 3 1 10 10 1 13 1 13 1320 1956 1961 1977 21 8 1979 1981 1982 2011 2003 20062011 Vol.9No.22012 119 1 23 8 21 1 1 24 20 22 18 8 120 Vol.9No.22012 2 9 1953 2 1979 2 2 1982
untitled
1 JAPAN TESTING ASSOCIATION 2 3 JAPAN TESTING ASSOCIATION 4 5 JAPAN TESTING ASSOCIATION 6 7 JAPAN TESTING ASSOCIATION 8 9 JAPAN TESTING ASSOCIATION 10 11 JAPAN TESTING ASSOCIATION 12 13 JAPAN TESTING ASSOCIATION
JOURNAL OF THE JAPAN FLUID POWER SYSTEM SOCIETY.. FLUID POWER SYSTEM
! JOURNAL OF THE JAPAN FLUID POWER SYSTEM SOCIETY.. FLUID POWER SYSTEM 267 268 269 270 271 !!!!!!!! 272 !!!!!!!!!!!!!!!!!!!! 273 274!!! 275! !!!! 276 277!!!!! 278 279 280 281 282! 283 !!! 284 Transient
PC Windows 95, Windows 98, Windows NT, Windows 2000, MS-DOS, UNIX CPU
1. 1.1. 1.2. 1 PC Windows 95, Windows 98, Windows NT, Windows 2000, MS-DOS, UNIX CPU 2. 2.1. 2 1 2 C a b N: PC BC c 3C ac b 3 4 a F7 b Y c 6 5 a ctrl+f5) 4 2.2. main 2.3. main 2.4. 3 4 5 6 7 printf printf
連立1次方程式Ax=bの解法：公式にしたがって解くのは，計算量大
Common Lisp プログラミング入門 概要 Lisp は記号の構造的な表現である S 式を操作するインタープリタ方式を基調とするプログラミング言語である. ここでは, 思考のツールとしての Lisp を強調した解説を行う.. Lisp のしくみ Lisp で中心となるのは,S 式 (Symbolic Expression) と呼ばれる記号の構造的な表現である.Lisp ユーザはインタープリタを使って,S
4 (induction) (mathematical induction) P P(0) P(x) P(x+1) n P(n) 4.1 (inductive definition) A A (basis ) ( ) A (induction step ) A A (closure ) A clos
4 (induction) (mathematical induction) P P(0) P(x) P(x+1) n P(n) 4.1 (inductive definition) A A (basis ) ( ) A (induction step ) A A (closure ) A closure 81 3 A 3 A x A x + A A ( A. ) 3 closure A N 1,
Scheme Hygienic Macro stibear (@stibear1996) 1 Scheme Scheme Lisp Lisp Common Lisp Emacs Lisp Clojure Scheme 1 Lisp Lisp Lisp Lisp Homoiconicity Lisper 2 Common Lisp gensym Scheme Common Lisp Scheme Lisp-1
C 2 / 21 1 y = x 1.1 lagrange.c 1 / Laglange / 2 #include <stdio.h> 3 #include <math.h> 4 int main() 5 { 6 float x[10], y[10]; 7 float xx, pn, p; 8 in
![C 2 / 21 1 y = x 1.1 lagrange.c 1 / Laglange / 2 #include <stdio.h> 3 #include <math.h> 4 int main() 5 { 6 float x[10], y[10]; 7 float xx, pn, p; 8 in](/thumbs/88/114913715.jpg "C 2 / 21 1 y = x 1.1 lagrange.c 1 / Laglange / 2 #include <stdio.h> 3 #include <math.h> 4 int main() 5 { 6 float x[10], y[10]; 7 float xx, pn, p; 8 in")
C 1 / 21 C 2005 A * 1 2 1.1......................................... 2 1.2 *.......................................... 3 2 4 2.1.............................................. 4 2.2..............................................
(CC Attribution) Lisp 2.1 (Gauche )
http://www.flickr.com/photos/dust/3603580129/ (CC Attribution) Lisp 2.1 (Gauche ) 2 2000EY-Office 3 4 Lisp 5 New York The lisps Sammy Tunis flickr lisp http://www.flickr.com/photos/dust/3603580129/ (CC
橡J0021-1.PDF
1 1950 60 PE PP PVD C PVC FT-lR DS C XRF GC LC 1 HS GC 2 2 1 2 1 GC GC 80 5 3 3 200 2 PVC 2 3 2 2 4 JHS 100 5 10ml GC 5 JHS-100 3 3 1 JHS 100A QP 5000 GC MS 6 1 1 TC-WAX 0.25mm 60m 40 180 5 /min--- 280
(search: ) [1] ( ) 2 (linear search) (sequential search) 1
![(search: ) [1] ( ) 2 (linear search) (sequential search) 1](/thumbs/97/131825822.jpg "(search: ) [1] ( ) 2 (linear search) (sequential search) 1")
2005 11 14 1 1.1 2 1.2 (search:) [1] () 2 (linear search) (sequential search) 1 2.1 2.1.1 List 2-1(p.37) 1 1 13 n
新・明解Ｃ言語で学ぶアルゴリズムとデータ構造
第 1 章 基本的 1 n 141 1-1 三値 最大値 algorithm List 1-1 a, b, c max /* */ #include int main(void) { int a, b, c; int max; /* */ List 1-1 printf("\n"); printf("a"); scanf("%d", &a); printf("b"); scanf("%d",
kiso2-06.key
座席指定があります Linux を起動して下さい 第6回 計算機基礎実習II 計算機基礎実習II 2018 のウェブページか ら 以下の課題に自力で取り組んで下さい 第5回の復習課題(rev05) 第6回の基本課題(base06) 第5回課題の回答例 ex05-2.c 1. キーボードから整数値 a を入力すると a*a*a の値を出力することを繰り返すプログラムを作成しなさい 2. ただし 入力された
ohp08.dvi
19 8 ( ) 2019.4.20 1 (linked list) ( ) next ( 1) (head) (tail) ( ) top head tail head data next 1: 2 (2) NULL nil ( ) NULL ( NULL ) ( 1 ) (double linked list ) ( 2) 3 (3) head cur tail head cur prev data
org/ghc/ Windows Linux RPM 3.2 GHCi GHC gcc javac ghc GHCi(ghci) GHCi Prelude> GHCi :load file :l file :also file :a file :reload :r :type expr :t exp
3 Haskell Haskell Haskell 1. 2. 3. 4. 5. 1. 2. 3. 4. 5. 6. C Java 3.1 Haskell Haskell GHC (Glasgow Haskell Compiler 1 ) GHC Haskell GHC http://www.haskell. 1 Guarded Horn Clauses III - 1 org/ghc/ Windows
Netcommunity SYSTEM X7000 IPコードレス電話機 取扱説明書
4 5 6 7 8 9 . 4 DS 0 4 5 4 4 4 5 5 6 7 8 9 0 4 5 6 7 8 9 4 5 6 4 0 4 4 4 4 5 6 7 8 9 40 4 4 4 4 44 45 4 6 7 5 46 47 4 5 6 48 49 50 5 4 5 4 5 6 5 5 6 4 54 4 5 6 7 55 5 6 4 56 4 5 6 57 4 5 6 7 58 4
.A. D.S
1999-1- .A. D.S 1996 2001 1999-2- -3- 1 p.16 17 18 19 2-4- 1-5- 1~2 1~2 2 5 1 34 2 10 3 2.6 2.85 3.05 2.9 2.9 3.16 4 7 9 9 17 9 25 10 3 10 8 10 17 10 18 10 22 11 29-6- 1 p.1-7- p.5-8- p.9 10 12 13-9- 2
24 200902728 1 4 1.1......................... 4 1.2......................... 4 1.3......................... 5 1.4......................... 5 1.5........................... 5 1.6...................... 6
‚æ4›ñ
( ) ( ) ( ) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z a b c d e f g h i j k l m n o p q r s t u v w x y z 0 1 2 3 4 5 6 7 8 9 (OUS) 9 26 1 / 28 ( ) ( ) ( ) A B C D Z a b c d z 0 1 2 9 (OUS) 9
土壌環境行政の最新動向（環境省 水・大気環境局土壌環境課）
201022 1 18801970 19101970 19201960 1970-2 1975 1980 1986 1991 1994 3 1999 20022009 4 5 () () () () ( ( ) () 6 7 Ex Ex Ex 8 25 9 10 11 16619 123 12 13 14 5 18() 15 187 1811 16 17 3,000 2241 18 19 ( 50
syuryoku
248 24622 24 P.5 EX P.212 2 P271 5. P.534 P.690 P.690 P.690 P.690 P.691 P.691 P.691 P.702 P.702 P.702 P.702 1S 30% 3 1S 3% 1S 30% 3 1S 3% P.702 P.702 P.702 P.702 45 60 P.702 P.702 P.704 H17.12.22 H22.4.1
N N 1,, N 2 N N N N N 1,, N 2 N N N N N 1,, N 2 N N N 8 1 6 3 5 7 4 9 2 1 12 13 8 15 6 3 10 4 9 16 5 14 7 2 11 7 11 23 5 19 3 20 9 12 21 14 22 1 18 10 16 8 15 24 2 25 4 17 6 13 8 1 6 3 5 7 4 9 2 1 12 13
Microsoft PowerPoint - IntroAlgDs-05-2.ppt
アルゴリズムとデータ構造入門 2005 年 10 月 11 日 アルゴリズムとデータ構造入門 1. 手続きによる抽象の構築 1.1 プログラムの要素 奥乃 博 1. TUT Schemeが公開されました. Windowsは動きます. Linux, Cygwin はうまく行かず. 調査中. 2. 随意課題 7の追加 友人の勉学を助け,TAの手伝いをする. 支援した内容を毎回のレポート等で詳細に報告.
卒 業 研 究 報 告.PDF
C 13 2 9 1 1-1. 1-2. 2 2-1. 2-2. 2-3. 2-4. 3 3-1. 3-2. 3-3. 3-4. 3-5. 3-5-1. 3-5-2. 3-6. 3-6-1. 3-6-2. 4 5 6 7-1 - 1 1 1-1. 1-2. ++ Lisp Pascal Java Purl HTML Windows - 2-2 2 2-1. 1972 D.M. (Dennis M Ritchie)
10/ / /30 3. ( ) 11/ 6 4. UNIX + C socket 11/13 5. ( ) C 11/20 6. http, CGI Perl 11/27 7. ( ) Perl 12/ 4 8. Windows Winsock 12/11 9. JAV
tutimura@mist.i.u-tokyo.ac.jp kaneko@ipl.t.u-tokyo.ac.jp http://www.misojiro.t.u-tokyo.ac.jp/ tutimura/sem3/ 2002 12 11 p.1/33 10/16 1. 10/23 2. 10/30 3. ( ) 11/ 6 4. UNIX + C socket 11/13 5. ( ) C 11/20
BW BW
Induced Sorting BW 11T2042B 2015 3 23 1 1 1.1................................ 1 1.2................................... 1 2 BW 1 2.1..................................... 2 2.2 BW.................................
28 27 8 4 10 17 2 27 8 7 14 00 1 27 8 14 15 00 2 27 8 21 15 00 1 4 5 2 6 1 27 ABCD 6 2 2 5 5 8% 108 100 49 2 13 140 22 12 7 153-8501 19 23 03-5478-1225 27 8 4 (1) (2) (3) (1) (2) (3) (4) (5) (6) (7) (8)
1 NScripter 1 [ NScripter ] NScripter NScripter 2 nathki bugyo 1 http://www.shuwasystem.co.jp/cgi-bin/detail.cgi?isbn=4-7980-1104-5 2 http://www.pulltop.com/gp04/ 2 NScripter NScripter BASIC ( ) NScLisper
BIT -2-
2004.3.31 10 11 12-1- BIT -2- -3-256 258 932 524 585 -4- -5- A B A B AB A B A B C AB A B AB AB AB AB -6- -7- A B -8- -9- -10- mm -11- fax -12- -13- -14- -15- s58.10.1 1255 4.2 30.10-16- -17- -18- -19-6.12.10
DOPRI5.dvi
ODE DOPRI5 ( ) 16 3 31 Runge Kutta Dormand Prince 5(4) [1, pp. 178 179] DOPRI5 http://www.unige.ch/math/folks/hairer/software.html Fortran C C++ [3, pp.51 56] DOPRI5 C cprog.tar % tar xvf cprog.tar cprog/
j x j j j + 1 l j l j = x j+1 x j, n x n x 1 = n 1 l j j=1 H j j + 1 l j l j E
8 9 7 6 4 2 3 5 1 j x j j j + 1 l j l j = x j+1 x j, n x n x 1 = n 1 l j j=1 H j j + 1 l j l j E a n 1 H = ae l j, j=1 l j = x j+1 x j, x n x 1 = n 1 j=1 l j, l j = ±l l > 0) n 1 H = ϵ l j, j=1 ϵ e x x
c-all.dvi
III(994) (994) from PSL (9947) & (9922) c (99,992,994,996) () () 2 3 4 (2) 2 Euler 22 23 Euler 24 (3) 3 32 33 34 35 Poisson (4) 4 (5) 5 52 ( ) 2 Turbo 2 d 2 y=dx 2 = y y = a sin x + b cos x x = y = Fortran
a-b...
,,.... a-b... m m, RC..... a-b, ....,. a-b .......... GHQA.. .,., KK... PHP..... .... a VOL............. b a-b . .......... ,, ,,,........... .... ...... . ,....... ......... ........... ........ ,,kg
£Ã¥×¥í¥°¥é¥ß¥ó¥°ÆþÌç (2018) - Â裵²ó ¨¡ À©¸æ¹½Â¤¡§¾ò·ïʬ´ô ¨¡
(2018) 2018 5 17 0 0 if switch if if ( ) if ( 0) if ( ) if ( 0) if ( ) (0) if ( 0) if ( ) (0) ( ) ; if else if ( ) 1 else 2 if else ( 0) 1 if ( ) 1 else 2 if else ( 0) 1 if ( ) 1 else 2 (0) 2 if else
n 第1章 章立ての部分は、書式(PC入門大見出し)を使います
FORTRAN FORTRAN FORTRAN ) DO DO IF IF FORTRAN FORTRAN(FORmula TRANslator)1956 IBM FORTRAN IV FORTRAN77 Fortran90 FORTRAN77 FORTRAN FORTARN IF, DO C UNIX FORTRAN PASCAL COBOL PL/I BASIC Lisp PROLOG Lisp
kiso2-09.key
座席指定はありません 計算機基礎実習II 2018 のウェブページか 第9回 ら 以下の課題に自力で取り組んで下さい 計算機基礎実習II 第7回の復習課題(rev07) 第9回の基本課題(base09) 第8回試験の結果 中間試験に関するコメント コンパイルできない不完全なプログラムなど プログラミングに慣れていない あるいは複雑な問題は 要件 をバラして段階的にプログラムを作成する exam08-2.c
OHP.dvi
0 7 4 0000 5.. 3. 4. 5. 0 0 00 Gauss PC 0 Gauss 3 Gauss Gauss 3 4 4 4 4 3 4 4 4 4 3 4 4 4 4 3 4 4 4 4 u [] u [3] u [4] u [4] P 0 = P 0 (),3,4 (,), (3,), (4,) 0,,,3,4 3 3 3 3 4 4 4 4 0 3 6 6 0 6 3 6 0 6
¥×¥í¥°¥é¥ß¥ó¥°±é½¬I Exercise on Programming I [1zh] ` `%%%`#`&12_`__~~~ alse
![¥×¥í¥°¥é¥ß¥ó¥°±é½¬I Exercise on Programming I [1zh] ` `%%%`#`&12_`__~~~ alse](/thumbs/97/134738565.jpg "¥×¥í¥°¥é¥ß¥ó¥°±é½¬I Exercise on Programming I [1zh] ` `%%%`#`&12_`__~~~ alse")
I Exercise on Programming I http://bit.ly/oitprog1 1, 2 of 14 ( RD S ) I 1, 2 of 14 1 / 44 Ruby Ruby ( RD S ) I 1, 2 of 14 2 / 44 7 5 9 2 9 3 3 2 6 5 1 3 2 5 6 4 7 8 4 5 2 7 9 6 4 7 1 3 ( RD S ) I 1, 2
01_教職員.indd
T. A. H. A. K. A. R. I. K. O. S. O. Y. O. M. K. Y. K. G. K. R. S. A. S. M. S. R. S. M. S. I. S. T. S. K.T. R. T. R. T. S. T. S. T. A. T. A. D. T. N. N. N. Y. N. S. N. S. H. R. H. W. H. T. H. K. M. K. M.
Microsoft PowerPoint - IntroAlgDs pptx
アルゴリズムとデータ構造入門 -14 2012 年 1 月 8 日 大学院情報学研究科知能情報学専攻 http://winnie.kuis.kyoto-u.ac.jp/~okuno/lecture/11/introalgds/ okuno@i.kyoto-u.ac.jp,okuno@nue.org if mod( 学籍番号の下 3 桁,3) 0 if mod( 学籍番号の下 3 桁,3) 1 if
Microsoft PowerPoint - IntroAlgDs-05-4.ppt
アルゴリズムとデータ構造入門 2005 年 0 月 25 日 アルゴリズムとデータ構造入門. 手続きによる抽象の構築.2 Procedures and the Processes They generate ( 手続きとそれが生成するプロセス ) 奥乃 博. TUT Scheme が公開されました. Windows は動きます. Linux, Cygwin も動きます. 0 月 25 日 本日のメニュー.2.
2 章 1 整数を一つ読み込み, その階乗を計算する RAM プログラムを書け f (n) = n! ( n 0) 何でもよい ( n 18 C ( ) hello world.c 1 #include <stdio.h> 2 3 main() 4 { 5 printf("hello World\n"); 6 } [ ] [ ] #include <stdio.h> % cc hello_world.c %./a.o
![18 C ( ) hello world.c 1 #include <stdio.h> 2 3 main() 4 { 5 printf(hello World\n); 6 } [ ] [ ] #include <stdio.h> % cc hello_world.c %./a.o](/thumbs/95/124976715.jpg "18 C ( ) hello world.c 1 #include <stdio.h> 2 3 main() 4 { 5 printf(hello World\n); 6 } [ ] [ ] #include <stdio.h> % cc hello_world.c %./a.o")
18 C ( ) 1 1 1.1 hello world.c 5 printf("hello World\n"); 6 } [ ] [ ] #include % cc hello_world.c %./a.out Hello World [a.out ] % cc hello_world.c -o hello_world [ ( ) ] (K&R 4.1.1) #include
ex12.dvi
1 0. C, char., char, 0,. C, ("),., char str[]="abc" ; str abc.,, str 4. str 3. char str[10]="abc" ;, str 10, str 3., char s[]="abc", t[10] ;, t = s. ASCII, 0x00 0x7F, char., "abc" 3, 1. 1 8 256, 2., 2
解きながら学ぶＣ言語
printf 2-5 37 52 537 52 printf("%d\n", 5 + 37); 5370 source program source file.c ex00.c 0 comment %d d 0 decimal -2 -p.6 3-2 5 37 5 37-22 537 537-22 printf("537%d\n", 5-37); function function call ( )argument,
Krylov (b) x k+1 := x k + α k p k (c) r k+1 := r k α k Ap k ( := b Ax k+1 ) (d) β k := r k r k 2 2 (e) : r k 2 / r 0 2 < ε R (f) p k+1 :=
127 10 Krylov Krylov (Conjugate-Gradient (CG ), Krylov ) MPIBNCpack 10.1 CG (Conjugate-Gradient CG ) A R n n a 11 a 12 a 1n a 21 a 22 a 2n A T = =... a n1 a n2 a nn n a 11 a 21 a n1 a 12 a 22 a n2 = A...
