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- ありみち みねむら
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1 I 2 tutimura@mist.i.u-tokyo.ac.jp tutimura/ I 2 p.1/??
2 / / Makefile I 2 p.2/??
3 Makefile make GNU make I 2 p.3/??
4 Makefile L A T E X I 2 p.4/??
5 core (1) gcc, gdb -g -ggdb % gcc hogehoge.c -O2 -Wall -g %./a.out Segmentation fault core sh/bash ulimit -c, csh/tcsh limit -h core I 2 p.5/??
6 core (2) % gdb./a.out core -g/-ggdb list where print quit I 2 p.6/??
7 Java... try {... catch (Exception e) { e.printstacktrace(); I 2 p.7/??
8 UNIX OS CTRL+D (shell) file command < file command > file command >> file (sh/bash) command 2> file (csh/tcsh) command >& file I 2 p.8/??
9 1 1 fopen()/fclose() fprintf(stderr, "..") C C++ Fortran Java stdin cin 5 System.in stdout cout 6 System.out stderr cerr System.err clog 0 I 2 p.9/??
10 I 2 p.10/??
11 % command1 command2 command1 command2 % command1 > tempfile % command2 < tempfile I 2 p.11/??
12 command less command sort command tee outputfile gzip -cd input.gz command command gzip > output.gz I 2 p.12/??
13 / (1) UNIX tar.gz hogehoge-1.0.tar.gz hogehoge-1.0/readme tar cvf - hogehoge-1.0/ gzip > hogehoge-1.0.tar.gz gzip -cd hogehoge-1.0.tar.gz tar tvf - gzip -cd hogehoge-1.0.tar.gz tar xvf - tar tar czvf hogehoge-1.0.tar.gz hogehoge-1.0/ tar tzvf hogehoge-1.0.tar.gz tar xzvf hogehoge-1.0.tar.gz I 2 p.13/??
14 / (2) MS-Windows zip/lzh hogehoge-1.0.zip README zip -r../hogehoge-1.0.zip * unzip -l hogehoge-1.0.zip unzip hogehoge-1.0.zip Makefile I 2 p.14/??
15 diff patch 2 -r -c, -u patch diff -ur software /memory_leak.c software /memory_leak.c --- software /memory_leak.c Sun Apr 21 12:29: software /memory_leak.c Wed Apr 24 22:52:46 -7,7 int *x; x = malloc(sizeof(x[0]) * 100); - for (i=0; i<101; i++) { + for (i=0; i<1001; i++) { x[i] = 0; I 2 p.15/??
16 C K&R (1973), void* ANSI (1989), void* C99 (1999) //, I 2 p.16/??
17 ... I 2 p.17/??
18 -III :2 [x 1, x 2 ] [y 1, y 2 ] [x 3, x 4 ] [y 3, y 4 ] (( )>0) int is_overlapped(int x1, int x2, int y1, int y2, int x3, int x4, int y3, int y4) { is_overlapped(0,5, 0,5, 2,6, 2,6) ==> TRUE is_overlapped(0,5, 0,5, 6,9, 6,9) ==> FALSE is_overlapped(0,5, 0,5, 0,5, 5,9) ==> FALSE I 2 p.18/??
19 x y x y int is_overlapped(int x1, int x2, int y1, int y2, int x3, int x4, int y3, int y4) { return is_overlapped2( x1, x2, x3, x4 ) && is_overlapped2( y1, y2, y3, y4 ); x y 2 I 2 p.19/??
20 x y I 2 p.20/??
21 /* (a,b) (c,d) */ int is_overlapped2(int a, int b, int c, int d) { a < b, c < d I 2 p.21/??
22 #include <assert.h> /* (a,b) (c,d) */ /* a < b, c < d */ int is_overlapped2(int a, int b, int c, int d) { assert( a < b ); assert( c < d ); return ( c < b && a < d ); a = b c = d I 2 p.22/??
23 #define TRUE 1 #define FALSE 0 #include <assert.h> /* (a,b) (c,d) */ /* a <= b, c <= d */ int is_overlapped2(int a, int b, int c, int d) { assert( a <= b ); assert( c <= d ); if ( a == b c == d ) return FALSE; return ( c < b && a < d ); I 2 p.23/??
24 #define TRUE 1 #define FALSE 0 #include <assert.h> /* (a,b) (c,d) (a<=b, c<=d) */ int is_overlapped(int a, int b, int c, int d) { assert( a <= b ); assert( c <= d ); if ( a == b c == d ) return FALSE; return ( c < b && a < d ); /* [x1,x2] [y1,y2] [x3,x4] [y3,y4] */ /* (x1<=x2, y1<=y2, x3<=x4, y3<=y4) */ int is_overlapped2(int x1, int x2, int y1, int y2, int x3, int x4, int y3, int y4) { return is_overlapped( x1, x2, x3, x4 ) && is_overlapped( y1, y2, y3, y4 ); I 2 p.24/??
25 #include <stdio.h> void test(int a, int b, int c, int d, int answer) { int ret = is_overlapped(a, b, c, d); if (ret &&!answer) { printf("[%d,%d] and [%d,%d] " "should not have any overlap.\n", a, b, c, d); else if (!ret && answer) { printf("[%d,%d] and [%d,%d] " "should have an overlap.\n", a, b, c, d); I 2 p.25/??
26 int main() { test(0,3, 4,9, FALSE); test(0,4, 3,8, TRUE); test(0,8, 3,4, TRUE); test(0,8, 3,8, TRUE); test(0,8, 0,8, TRUE); test(0,0, 0,0, FALSE); test(0,8, 1,1, FALSE); /* 0 */ test(0,1, 8,8, FALSE); test(0,0, 0,8, FALSE); test(0,4, 4,9, FALSE); return 0; I 2 p.26/??
27 (1) I 2 p.27/??
28 (2)... I 2 p.28/??
29 -IV [a, b] [c, d] /* [a,b] [c,d] */ int intersection(int a, int b, int c, int d) { intersection(1,5, 2,6) ==> 3 intersection(1,2, 3,4) ==> 0 I 2 p.29/??
30 #include <assert.h> #define MAX(a,b) ( ( (a)>=(b) )? (a):(b) ) /* */ #define MIN(a,b) ( ( (a)<=(b) )? (a):(b) ) /* */ /* [a,b] [c,d] a<=b,c<=d */ int intersection(int a, int b, int c, int d) { int lower, upper, length; assert(a <= b && c <= d); lower = MAX(a,c); upper = MIN(b,d); length = upper - lower; if (length < 0) return 0; return length; I 2 p.30/??
31 #include <assert.h> /* */ /* [a,b] [c,d] a<=b,c<=d */ int intersection(int a, int b, int c, int d) { int i, j, k; assert(a <= b && c <= d); if (a > c) i = a; else i = c; /* i = max(a,c); */ if (b > d) j = d; else j = b; /* j = min(b,d); */ k = i - j; if (k < 0) return 0; return k; I 2 p.31/??
32 /* [a,b] [c,d] a<=b,c<=d */ int intersection_normalized(int a, int b, int c, int d) { /* [a,b] [c,d] [2,1] */ int intersection(int a, int b, int c, int d) { if ( a > b ) { fprintf(stderr, "intersection: a > b\n"); exit(1); if ( c > d ) { fprintf(stderr, "intersection: c > d\n"); exit(1); return intersection_normalized(a,b,c,d); /* [a,b] [c,d] [2,1] 0 */ int intersection2(int a, int b, int c, int d) { if ( a > b ) return 0; if ( c > d ) return 0; return intersection_normalized(a,b,c,d); I 2 p.32/??
33 void test(int a, int b, int c, int d, int answer) { int ret = intersection(a, b, c, d); if (ret!= answer) { printf("error! [%d,%d] [%d,%d] ==> %d, which should be %d.\n", a, b, c, d, ret, answer); int main() { test(1,5, 2,6, 3); test(5,1, 6,2, 3); test(1,2, 3,4, 0); test(1,1, 1,1, 0); test(0,0, 0,0, 0); test(-1,-1, -1,-1, 0); /* 1, 2, 5, 6 */ test(1,2, 5,6, 0); I 2 p.33/??
34 a, b, c, d a, b, c, d X, Y, Z d!d b!b a!a a!a c X Y X Y!c Y X Y X c X X X X!c Z Z Z Z : if (a && b && c && d) { X(); else if (!a && b && c && d) { Y();... I 2 p.34/??
35 (1) if ( d ) { if ( (a && c) (!a &&!c) ) { X(); else { Y(); else { if ( c ) { X(); else { Z(); d!d b!b a!a a!a c X Y X Y!c Y X Y X c X X X X!c Z Z Z Z I 2 p.35/??
36 (2) #define XOR(a,b) (!(a) ˆ!(b) ) /* (!(a)!=!(b) ) */ if (!d &&!c ) Z(); else if (!d && c ) X(); else if ( XOR(a,c) ) Y(); else X(); d!d b!b a!a a!a c X Y X Y!c Y X Y X c X X X X!c Z Z Z Z I 2 p.36/??
37 (3) int table[] = { 1,2,1,2, 2,1,2,1, 1,1,1,1, 3,3,3,3 ; int index = 0; if (!a) index += 1; if (!b) index += 2; if (!c) index += 4; if (!d) index += 8; switch (table[index]) { case 1: X(); break; case 2: Y(); break; case 3: Z(); break; d b!b a!a a!a c X 0 Y 1 X 2 Y 3!c Y 4 X 5 Y 6 X 7!d c X8 X 9 X 10 X 11!c Z 12 Z 13 Z 14 Z 15 I 2 p.37/??
38 (4) void (*func[])() = { X, Y, X, Y, Y, X, Y, X, X, X, X, X, Z, Z, Z, Z ; int index = 0; if (!a) index += 1; if (!b) index += 2; if (!c) index += 4; if (!d) index += 8; (*func[index])(); d!d b!b a!a a!a c X Y X Y!c Y X Y X c X X X X!c Z Z Z Z I 2 p.38/??
39 1 1 I 2 p.39/??
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ohp03.dvi
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0. 目次 6. 2 項係数 7. 二分探索 8. 最大値探索 9. 集合 {1,2,,n} 上の部分集合生成 - 1 - 6. 2 項係数 再帰的定義 2 項係数 c(n,r) は つぎのように 定義される c(n,r) = c(n-1,r) + c(n-1,r-1) (n 2,1 r n-1) = 1 (n 0, r=0 ) = 1 (n 1, r=n ) c(n,r) 0 1 2 3 4 5
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ex01.dvi
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Informatics 2015
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