(2018) 2018 5 24
( ) while ( ) do while ( ); for ( ; ; )
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++;
while int i = 0; i while (i < 100) { printf("i = %3d\n", i); i++;
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; 2 0<100
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; 2 3
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; i i=1 2 3 4 i 1
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; 2 3 4 i 1 2 4
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; 2 3 4 i 1 1<100 2 4
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; 2 3 4 i 1 2 4
while int i = 0; while (i < 100) { printf("i = %3d\n", i); i++; i i=2 2 3 4 i 1 2 4
while int i = 0; while (i < 100) { 100>=100 printf("i = %3d\n", i); i++; 2 3 4 i 1 2 4 2
ex15c int main(int argc, char **argv) { int i = 10, imax; printf("imax = "); scanf("%d", &imax); while (i <= imax) { printf("i = %3d\t i**2 = %5d\n", i, i * i); i++; return 0;
ex15c int main(int argc, char **argv) { int i = 10, imax; printf("imax = "); scanf("%d", &imax); while (i <= imax) { printf("i = %3d\t i**2 = %5d\n", i, i * i); i++; gcc ex15c -o ex15 -lm -Wall return 0;
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ;
do while int i = 0; i do { printf("i = %3d\n", i); i++; while (i < 100 ) ;
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 2
do while int i = 0; do { printf("i = %3d\n", i); i++; i i=1 while (i < 100 ) ; 2 3 i 1
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 2 3 i 1 4 1<100
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 2 3 i 1 4 2 4
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 2 3 i 1 4 2<100 2 4
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 2 3 i 1 4 2 4
do while int i = 0; do { printf("i = %3d\n", i); i++; i i=3 while (i < 100 ) ; 2 3 i 1 4 2 4
do while int i = 0; do { printf("i = %3d\n", i); i++; while (i < 100 ) ; 100>=100 2 3 i 1 4 2 4 4
ex16c int main(int argc, char **argv) { int i = 10, imax; printf("imax = "); scanf("%d", &imax); do { printf("i = %3d\t i**2 = %5d\n", i, i * i); i++; while (i <= imax); return 0;
ex16c int main(int argc, char **argv) { int i = 10, imax; printf("imax = "); scanf("%d", &imax); do { printf("i = %3d\t i**2 = %5d\n", i, i * i); i++; while (i <= imax); gcc ex16c -o ex16 -lm -Wall return 0;
for for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ;
for for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; i
for 0<100 for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2
for for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2 3
for for (i = 0; i < 100; i++) { i i=1 printf("i = %3d\n", i) ; 2 3 4 i (1 )
for for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2 3 4 i (1 ) 2 4
for 1<100 for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2 3 4 i (1 ) 2 4
for for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2 3 4 i (1 ) 2 4
for for (i = 0; i < 100; i++) { i i=2 printf("i = %3d\n", i) ; 2 3 4 i (1 ) 2 4
for 100>=100 for (i = 0; i < 100; i++) { printf("i = %3d\n", i) ; 2 3 4 i (1 ) 2 4 2
ex17c int main(int argc, char **argv) { int i, imax; printf("imax = "); scanf("%d", &imax); for (i = 10; i <= imax; i++) { printf("i = %3d\t i**2 = %5d\n", i, i * i); return 0;
ex17c int main(int argc, char **argv) { int i, imax; printf("imax = "); scanf("%d", &imax); for (i = 10; i <= imax; i++) { printf("i = %3d\t i**2 = %5d\n", i, i * i); gcc ex17c -o ex17 -lm -Wall return 0;
ex18c int main(int argc, char **argv) { int i, j, n; for (i = 0; i < 16; i++) { for (j = 0; j < 8; j++) { n = 8 * i + j; printf("%3d(x%02x) ", n, n); printf("\n"); return 0;
return break switch continue goto
ex19c int main(int argc, char **argv) { char c; while (1) { printf("input a character "); scanf(" %c", &c); if (c == q c == Q ) break; printf("%c\n", c); printf("now quit\n"); return 0;
x ±1 ++x x++ - -x x- -
x ±1 ++x x++ - -x x- - ++ 1 - - 1 ++x - -x x++ x- -
==!= >= <= > <
2 p, q 2 x 2 + px + q = 0 1 2 2 15 2 p ) 1Y17B023 p = 23 3 q q = 10 10 12 4 "Imaginary roots"
$ /hw2018-1 p, q = 101, 10 1010000e+02, 1000000e+01 Solution1: xp = -00991071507656756, xm = -100900892849234 Solution2: xp = -00991071507656722, xm = -100900892849234 $ /hw2018-1 p, q = 101, 1e-12 1010000e+02, 1000000e-12 Solution1: xp = -7105427357601e-15, xm = -101 Solution2: xp = -99009900990099e-15, xm = -101 $ /hw2018-1 p, q = 101,2600 1010000e+02, 2600000e+03 Imaginary roots
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