第5章（第四版）C语言程序设计练习
一、例题
1. 1+2+3+···+100，即Σ100 n=1 n
#include<stdio.h> int main() { int i = 1; int sum = 0; while (i<=100) { sum = sum + i; i++; } printf("%d", sum); return 0; }
2.do...while
#include<stdio.h> int main() { int i = 1; int sum = 0; do { sum = sum + i; i++; } while (i <= 100); printf("%d", sum); return 0; }
3.
4.1000名学生，募捐达到10万元结束，统计捐款人数和数目
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #define SUM 100000 int main() { float amount,aver,sum; int i; for (i = 1,sum=0; i <= 1000; i++) { printf("请输入"); scanf("%f", &amount); sum = sum + amount; if (sum >= SUM) break; } aver = sum / i; printf("aver=%f\nnum=%d", aver, i); return 0; }
5.100~200之间，不能被3整除的数
#include<stdio.h> int main() { int i = 0; for (i = 100; i <= 200; i++) { if (0 == i % 3) continue; printf("%d ", i); } return 0; }
6.输出4*5矩阵
1 2 3 4 5
2 4 6 8 10
3 6 9 12 15
4 8 12 16 20
#include<stdio.h> int main() { int i, j, n = 0; for (i = 1; i <= 4; i++) for (j = 1; j <= 5; j++,n++) { if (n % 5 == 0) printf("\n"); printf("%d\t", i * j); } return 0; }
7.∏/4 = 1- 1⁄3 +1⁄5 - 1⁄7 + ···
Π ≈ 22⁄7
Π2 ⁄6 ≈ 1⁄1 + 1⁄4 + 1⁄9 ···+ 1⁄n*n
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<math.h> int main() { int sign = 1, count = 0; double pi = 0; double n = 1; double term = 1; while (fabs(sign/n)>=1e-6)//1乘以10的负6次方 { pi = pi + sign / n; n = n + 2; sign = -sign; count++; } pi = pi * 4; printf("pi=%lf\n", pi); printf("count=%d\n", count); return 0; }
8.Fibonacci
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> //通过公式计算，效率低 //造成大量重复计算 int count=0; int Fib(int n) { if (n == 3) { count++; } if (n <= 2) return 1; else return Fib(n - 1) + Fib(n - 2); } int main() { int n = 0; int ret = 0; scanf("%d", &n); ret = Fib(n); printf("ret=%d\n", ret); printf("count=%d\n",count); return 0; }
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> //循环或迭代的方式 //倒着进行计算 int Fib(int n) { int a = 1; int b = 1; int c = 1; while (n > 2) { c = a + b; //产生新的a、b，进行赋值 a = b; b = c; n--; } return c;//n为1或2，这里执行，返回c=1 } int main() { int n = 0; int ret = 0; scanf("%d", &n); ret = Fib(n); printf("ret=%d\n", ret); //printf("count=%d\n",count); return 0; }
9.判断素数
#define _CRT_SECURE_NO_WARNINGS #include<stdio.h> int main() { int n; int i; printf("请输入大于3的整数："); scanf("%d", &n); for (i = 2; i <= n-1; i++) if (n % i == 0) break; if (i < n) printf("是素数"); else printf("不是素数"); return 0; }
10.100~200之间全部素数
#include<stdio.h> #include<math.h> int prime(int n) { //2~n-1 之间的数字 int j = 0; for (j = 2; j <=sqrt(n); j++) { if (n % j == 0)//即n被整除，不是素数 return 0;//返回0 } return 1;//返回整数 } int main() { //100-200 int i = 0; int count=0; for (i = 100; i <= 200; i++) { //判断i是否为素数 if (prime(i) == 1) { count++ ； printf(" %d", i ); } } printf(" \ncount = %d\n",count); return 0; }
11.电文变成密码： 将字母A变为E、将字母a变为e，即变成其后的第四个字母，W将变成A。字母按上述规律转换，非字母字符不变。输入一行字符，输出相应密码
#define _CRT_SECURE_NO_WARNINGS //#include <stdio.h> //void main() //{ // char c; // while ((c = getchar()) != '\n') // { // if ((c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z')) // { // c = c + 4; // if (c > 'Z' && c <= 'Z' + 4 || c > 'z') // c = c - 26; // } // printf("%c", c); // } // printf("\n"); //} #include<stdio.h> int main() { char ch; while ((ch = getchar()) != '\n') { if ((ch >= 'A' && ch <= 'V') || (ch >= 'a' && ch <= 'v')) { ch = ch + 4; } if (ch >= 'W' && ch <= 'Z') { ch = ch - 26; } printf("%c",ch); } printf("\n"); return 0; }
二、习题
3.
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> int main() { int p, r, n, m, tmp; printf("请输入两个正整数n,m:"); scanf("%d,%d,", &n, &m);//5,3 if (n < m) { tmp = n; n = m; m = tmp; } p = n * m;//15 while (m != 0) { r = n % m;//5%3=2;3%2=1;2%1=0 n = m;//3;2;1 m = r;//2;1;0 } printf("它们的最大公约数为:%d\n", n);//1 printf("它们的最小公倍数为:%d\n", p / n);//15/1 return 0; }
4.
#include <stdio.h> int main() { char c; int letters = 0, space = 0, digit = 0, other = 0; printf("请输入一行字符:\n"); while ((c = getchar()) != '\n') { if (c >= 'a' && c <= 'z' || c >= 'A' && c <= 'Z') letters++; else if (c == ' ') space++; else if (c >= '0' && c <= '9') digit++; else other++; } printf("字母数:%d\n空格数:%d\n数字数:%d\n其它字符数:%d\n", letters, space, digit, other); return 0; }
5.
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> int main() { int a, n, i = 1, sn = 0, tn = 0; printf("a,n=:"); scanf("%d,%d", &a, &n);//1,2 while (i <= n)//1<2;2=2 { tn = tn + a; //0+1;1+10=11 sn = sn + tn; //0+1;11+1=12 a = a * 10;//10;100 ++i;//2;3 } printf("a+aa+aaa+...=%d\n", sn); return 0; }
6.
#include <stdio.h> int main() { long long n = 1, sum = 0; int i = 1; for (i = 1; i <= 20; i++) { n *= i; sum += n; } printf("%I64d", sum);//2561327494111820313 //printf("%lld", sum); }
7.
#include <stdio.h> int main() { int n1 = 100, n2 = 50, n3 = 10; double k, s1 = 0, s2 = 0, s3 = 0; for (k = 1; k <= n1; k++) { s1 = s1 + k; } for (k = 1; k <= n2; k++) { s2 = s2 + k * k; } for (k = 1; k <= n3; k++) { s3 = s3 + 1 / k; } printf("%f\n", s1); printf("%f\n", s2); printf("%f\n", s3); printf("sum=%15.6f\n", s1 + s2 + s3); return 0; }
8.
#include <stdio.h> int main() { int i, j, k, n; for (n = 100; n < 1000; n++) { //153 i = n / 100;//1 j = n / 10 - i * 10;//15-10=5 k = n % 10;//153%10=3 if (n == i * i * i + j * j * j + k * k * k) printf("%d ", n); } printf("\n"); return 0; }
9.
#include <stdio.h> int main() { int m, s, i; for (m = 2; m < 1000; m++) { //m=6 s = 0; for (i = 1; i < m; i++)//1;2;3 if ((m % i) == 0) //0;0;0 s = s + i;//1;3;6 if (s == m) { printf("%d,its factors are ", m); for (i = 1; i < m; i++)//1;2;3 if (m % i == 0) printf("%d ", i); printf("\n"); } } return 0; }
10.
#include <stdio.h> int main() { int i, n = 20; double a = 2, b = 1, s = 0, t; for (i = 1; i <= n; i++) { s = s + a / b;//2/1;2/1+3/2 t = a;//2;3; a = a + b;//3;5 b = t;//2;3 } printf("sum=%16.10f\n", s); return 0; }
11
#include <stdio.h> int main() { double sn = 100, hn = sn / 2; int n; for (n = 2; n <= 10; n++) { sn = sn + 2 * hn; hn = hn / 2; } printf("第10次落地时共经过%f米\n", sn); printf("第10次反弹%f米\n", hn); return 0; }
12
#include <stdio.h> int main() { int day, x1, x2;//x1为第一天桃子个数 day = 9; x2 = 1; while (day > 0) { x1 = (x2 + 1) * 2; x2 = x1; day--; } printf("total=%d\n", x1);//1534 return 0; }
13
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <math.h> int main() { float a, x0, x1; printf("enter a positive number:"); scanf("%f", &a); x0 = a / 2; x1 = (x0 + a / x0) / 2; do { x0 = x1; x1 = (x0 + a / x0) / 2; } while (fabs(x0 - x1) >= 1e-5); printf("The square root of %5.2f is %8.5f\n", a, x1); return 0; }
14
#include <stdio.h> #include <math.h> int main() { double x1, x0, f, f1; x1 = 1.5; 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) >= 1e-5); printf("The root of equation is %5.2f\n", x1); return 0; }
15
#define _CRT_SECURE_NO_WARNINGS #include <stdio.h> #include <math.h> int main() { float x0, x1, x2, fx0, fx1, fx2; do { printf("enter x1 & x2:"); scanf("%f,%f", &x1, &x2); fx1 = x1 * ((2 * x1 - 4) * x1 + 3) - 6; fx2 = x2 * ((2 * x2 - 4) * x2 + 3) - 6; } while (fx1 * fx2 > 0); do { x0 = (x1 + x2) / 2; fx0 = x0 * ((2 * x0 - 4) * x0 + 3) - 6; if ((fx0 * fx1) < 0) { x2 = x0; fx2 = fx0; } else { x1 = x0; fx1 = fx0; } } while (fabs(fx0) >= 1e-5); printf("x=%6.2f\n", x0); return 0; }
16
#include <stdio.h> int main() { int i, j, k; for (i = 0; i <= 3; i++) { for (j = 0; j <= 2 - i; j++) printf(" "); for (k = 0; k <= 2 * i; k++) printf("*"); printf("\n"); } for (i = 0; i <= 2; i++) { for (j = 0; j <= i; j++) printf(" "); for (k = 0; k <= 4 - 2 * i; k++) printf("*"); printf("\n"); } return 0; }
17
#include <stdio.h> int main() { char i, j, k; for (i = 'x'; i <= 'z'; i++) for (j = 'x'; j <= 'z'; j++) if (i != j) for (k = 'x'; k <= 'z'; k++) if (i != k && j != k) if (i != 'x' && k != 'x' && k != 'z') printf("A--%c\nB--%c\nC--%c\n", i, j, k);//A--z;B--x;C--y return 0; }
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