例5.1
#include#include int main() { int n = 13, year; double number, rate = 0.02; for (year = 1; year <= 10; year++) { number = n * pow((1 + rate), year); printf("%2d年后,人数为:%.2f亿n", year, number); } return 0; }
例5.2
#include#include int main(void) { double sum, item, flag, denominator; sum = 0; item = 1; flag = 1; denominator = 1; while (fabs(item) >= 1e-6) { sum = sum + item; flag = -flag; denominator = denominator + 3; item = flag / denominator; } printf("sum=%fn",sum); return 0; }
5.2.1 for语句的基本语法
for 语句的一般形式为：
for(表达式1;表达式2;表达式 3）
循环体语句：
for 语句的执行过程如下：
①首先计算表达式 1
②判断表达式 2,若其值为真(非0)，则执行循环体语句,然后
执行第③步;若值为假(0)，结束循环,转到第⑤步执行。
③ 计算表达式 3。
④返回第②步继续执行。
⑤ 循环结束,继续执行 for 语句的下一条语句。
大部分情况下,循环体语句为一复合语句。
例5.3
#includeint main(void) { int i, n, sum; scanf_s("%d", &n); sum = 0; for (i = 1; i <= n; i++) sum = sum + i; printf("由1到%d的和是:%dn",n,sum); return 0; }
例5.4
#includeint main(void) { int i, n; double factorial; printf("输入n的值:"); scanf_s("%d", &n); factorial = 1; for (i = 1; i <= n; i++) factorial = factorial * i; printf("%d!=%.0fn", n, factorial); return 0; }
例5.5
例5.6
#includeint main(void) { int i; float x,max; printf("请输入第1个数:"); scanf_s("%f", &x); max = x; for (i = 1; i <= 9; i++) { printf("请输入第%d个数:", i + 1); scanf_s("%f", &x); if (x > max) max = x; } printf("10个数的最大值是:%.0f",max); return 0; }
例5.7
#includeint main(void) { int number, a, b, c; for (number = 100; number <= 999; number++) { a = number / 100; b = number % 100 / 10; c = number % 10; if (number == a * a * a + b * b * b + c * c * c) printf("%5d", number); } return 0; }
例5.8
#includeint main(void) { int number, sum,i; printf("请输入一个正整数:"); scanf_s("%d", &number); sum = 0; for (i = 1; i <= number - 1; i++) if (number % i == 0) sum = sum + i; if (number == sum) printf("%d是完数n", number); else printf("%d不是完数n", number); return 0; }
例5.9
#includeint main(void) { int upper,lower,digit,i,other; char ch; upper = lower = digit = other = 0; printf("输入10个字符:"); for (i = 1; i <= 10; i++) { ch = getchar(); if (ch >= 'a' && ch <= 'z') lower++; else if (ch >= 'A' && ch <= 'Z') upper++; else if (ch >= '0' && ch <= '9') digit++; else other++; } printf("小写字母%d个,大写字母%d个,数字%d个,其他字符%d个n",lower,upper, digit, other); return 0; }
例5.10
#includeint main(void) { int i; char ch; for (i = 1; (ch = getchar()) != 'n'; i++) putchar(ch - 32); return 0; }
例5.11
#includeint main(void) { int n=0,i; char ch; printf("输入3个数字:"); for (i = 1; i <= 3; i++) { scanf_s("%c", &ch); n = n * 10 + ch - '0'; } printf("%dn", n); return 0; }
例5.12
#includeint main(void) { int i,flag,number; printf("请输入一个正整数:"); scanf_s("%d", &number); flag = 1; for (i = 2; i <= number - 1 && flag; i++) if (number % i == 0) flag = 0; if (flag) printf("%d是素数n", number); else printf("%d不是素数n", number); return 0; }
例5.13
#includeint main(void) { int i, sum; i = 1; sum = 0; while (i <= 100) { sum = sum + i; i = i + 1; } printf("sum=%dn",sum); return 0; }
例5.14
#includeint main(void) { int n=0,i; char ch; printf("输入3个数字:"); for (i = 1; i <= 3; i++) { scanf_s("%c", &ch); n = n * 10 + ch - '0'; } printf("%dn", n); return 0; }
例5.15
#includeint main(void) { int digit, letter, other = 0; char ch; digit = letter = other = 0; printf("请输入一串字符:"); while ((ch = getchar()) != 'n') if ((ch >= "0") && (ch <= '9')) digit++; else if ((ch >= 'a' && ch <= 'z') || (ch >= 'A' && ch <= 'Z')) letter++; else other++; printf("数字%d个,字母%d个,其他%d个n",digit,letter,other); return 0; }
例5.16
#includeint main(void) { int i, sum; i = 1; sum = 0; do { sum = sum + i; i = i + 1; } while (i <= 100); printf("sum=%dn", sum); return 0; }
例5.17
#includeint main(void) { int a, b, r, n, m; printf("请输入两个整数:"); scanf_s("%d%d", &a, &b); m = a, n = b; do { r = a % b; a = b; b = r; } while (r != 0); printf("%d和%d的最大公约数是:%dn",m,n,a); printf("最小公倍数是:%d", m * n / a); return 0; }
例5.18
#include5.5 改变循环结构的跳转语句int main(void) { long n, m; int count = 0; printf("请输入一个整数:"); scanf_s("%ld", &n); m = n; if (n < 0)n = -n; do { n = n / 10; count++; } while (n != 0); printf("整数%ld有%d位数n", m,count); return 0; }
例5.19
#includeint main(void) { int i = 5; do { if (i % 3 == 1) if (i % 5 == 2) { printf("%d", i); break; } i++; } while (i!= 0); return 0; }
例5.20
#include#include int main(void) { int n, m, i; printf("请输入一个正整数:"); scanf_s("%d", &n); m = sqrt(n); for (i = 2; i <= m; i++) if (n % i == 0) break; if (i > m) printf("%d是素数!n", n); else printf("%d不是素数!n", n); return 0; }
例5.21
#include5.5.2 continue 语句int main() { int num, n; float score, total = 0; num = 0; n = 0; while (1) { printf("请输入分数#%d(0~100):", n + 1); scanf_s("%f", &score); if (score < 0) break; if (score < 60) num++; total = total + score; n++; } printf("平均分数是:%.2f.n", total / n); printf("不及格的有:%d.n", num); return 0; }
例5.22
#includeint main() { int i, n = 1; for (i = 1; i <= 100; i++) { if (i % 7 != 0) continue; printf("%4d", i); if (n++ % 5 == 0)printf("n"); } return 0; }
例5.23
#includeint main() { int n, s = 0; n = 1; while (n < 10) { s = s + n; if (s > 5) break; if (n % 2 == 1) continue; n++; } printf("s=%d,n=%dn",s,n ); return 0; }
例5.24
#includeint main() { int i, sum; i = 1; sum = 0; loop:if (i <= 100) { sum = sum + i; i = i + 1; goto loop; } printf("sum=%dn", sum); return 0; }
5.6 循环嵌套
例5.25
#includeint main() { int i, j; double factorial, s = 0; for (i = 1; i <= 10; i++) { factorial = 1; for (j = 1; j <= i; j++) factorial = factorial * j; s = s + factorial; } printf("1!+2!+3!+...+10!=%.0fn",s); return 0; }
例5.26
#includeint main() { int i, j; for (i = 1; i <= 9; i++) { for (j = 1; j <= i; j++) printf("%d*%d=%dt", j, i, j * i); printf("n"); } return 0; }
例5.27
#include#include int main() { int i, n, k, count = 0; n = 2; while (n < 100) { k = sqrt(n); for (i = 2; i <= k; i++) if (n % i == 0)break; if (i > k) { printf("%4d", n); if (++count % 10 == 0)printf("n"); } n++; } return 0; }
例5.28
#include5.7 典型算法举例int main() { int i, n, m; for (m = 10; m <= 20; m++) { n = m, i = 2; printf("%d=",n); do { if (n % i == 0) { printf("%d*", i); n = n / i; } else i++; } while (n != i); printf("%dn", n); } return 0; }
例5.29
#includeint main() { int day, d1, d2; day = 9; d2 = 1; do { d1 = (d2 + 1) * 2; d2 = d1; --day; } while (day > 0); printf("第一天摘了%dn", d1); return 0; }
例5.30
#include#include #define eps 1e-6 int main() { int n = 1; float x; double fz, fm = 1, sinx; printf("输入x的值:"); scanf_s("%f", &x); fz = x; sinx = x; do { n = n + 1; fz = -fz * x * x; fm = fm * (2 * n - 2) * (2 * n - 1); sinx = sinx + fz / fm; } while (fabs(fz / fm) > eps); printf("sin(%f)=%0.6fn", x, sinx); printf("sin(%f)=%0.6fn", x, sin(x)); return 0; }
例5.31
#include#include #define eps 1e-6 int main() { float x1, x0, f, f1; x1 = 1.0; do { x0 = x1; f = ((2 * x0 - 4) * x0 + 3) * x0 - 6; f1 = (6 * x0 - 8) * x0 + 3; x1 = x0 - f / f1; } while (fabs(x1 - x0) >eps); printf("%6.2f", x1); return 0; }
例5.32
#includeint main() { int men, women, child; for(men=0;men<=9;men++) for (women = 0; women <= 12; women++) { child = 36 - men - women; if (men * 4 + women * 3 + child * 0.5 == 36) printf("男:%d,女:%d,小孩:%dn", men, women, child); } return 0; }
例5.33
#includeint main() { int i, j, k, n = 0; for(i=1;i<5;i++) for(j=1;j<5;j++) for(k=1;k<5;k++) if (i != k && i != j && j != k) { printf("%d%d%dt", i, j, k); if (++n % 5 == 0)printf("n"); } printf("n共有:%dn", n); return 0; }
例5.34
#includeint main() { int i, j; for (i = 1; i <= 5; i++) { for (j = 1; j <= 20 - i; j++) printf(" "); for (j = 1; j <= 2 * i - 1; j++) printf("*"); printf("n"); } return 0; }
例5.35
#include#include int main() { int m, n, count = 0; m = rand() % (80 - 10 + 1) + 10; printf("请输入一个10-80之间的整数:"); while (1) { scanf_s("%d", &n); count++; if (m == n) { printf("恭喜!你猜对了,你真棒!n"); break; } else if (m > n && count < 5) printf("对不起!你猜小了!再来一次!"); else if (m < n && count < 5) printf("对不起!你猜大了!再来一次!"); if (count == 5) { printf("对不起!你没有机会了!n这个数是:%d,游戏结束!n", m); break; } } return 0; }
例5.36
#include#include int main() { int x, i, j = 0, n, k = 0; for (x = 100; x < 1000; x++) { k = sqrt(x); for (i = 2; i <= k; i++) if (x % i == 0) break; if (i > k) { k = x; n = 0; while (k > 0) { n = n * 10 + k % 10; k /= 10; } if (x == n) { printf("%dt", x); if (++j % 5 == 0)printf("n"); } } } return 0; }
例5.37
#includeint main(void) { int x, t; printf("itpowern"); for (x = 100; x < 1000; x++) { t = x * x; while (t != 0) { if (x == t % 1000) { printf("%dt%dn", x, x * x); break; } else t = t / 10; } } return 0; }
