实验任务一
程序源码:
#include<stdio.h> #include<stdlib.h> #include<time.h> #include<windows.h> #define N 80 void print_text(int line,int col,char text[]);//函数声明 void print_spaces(int n);//函数声明 void print_blank_lines(int n);//函数声明 int main(){ int line,col,i; char text[N]="hi,April~"; srand(time(0));//以当前的系统时间作为随机种子 for(i=1;i<=10;++i){ line=rand()%25; col=rand()%80; print_text(line,col,text); Sleep(1000);//暂停1000ms } return 0; } //打印n个空格 void print_spaces(int n){ int i; for(i=1;i<=n;++i) printf(" "); } //打印n行空白行 void print_blank_lines(int n){ int i; for(i=1;i<=n;++i) printf("\n"); } //在第line行第col列打印一段文本 void print_text(int line,int col,char text[]){ print_blank_lines(line-1);//打印(line-1)行空行 print_spaces(col-1);//打印(col-1)列空格 printf("%s",text);//在第line行、col列输出text中字符串 }
程序运行:
问题回答:
每隔1秒输出一个“hi,April~”,共输出10个
实验任务二
task2.1
程序源码:
//利用局部static变量的特性,计算阶乘 #include<stdio.h> #include<stdlib.h> long long fac(int n);//函数声明 int main(){ int i,n; printf("Enter n:"); scanf("%d",&n); for(i=1;i<=n;++i) printf("%d!=%lld\n",i,fac(i)); system("pause"); return 0; } //函数定义 long long fac(int n){ static long long p=1; p=p*n; return p; }
程序运行:
task2.2
程序源码:
//练习:局部static变量特性 #include<stdio.h> int func(int,int); //函数声明 int main(){ int k=4,m=1,p1,p2; p1=func(k,m); //函数调用 p2=func(k,m); //函数调用 printf("%d,%d\n",p1,p2); return 0; } //函数定义 int func(int a,int b){ static int m=0,i=2; i+=m+1; m=i+a+b; return m; }
程序运行:
实验任务三
程序源码:
#include<stdio.h> long long func(int n);//函数声明 int main() { int n; long long f; while (scanf_s("%d", &n) != EOF) { f = func(n);//函数调用 printf("n=%d,f=%lld\n", n, f); } return 0; } //函数定义 long long func(int n) { if (n == 1) return 1; if (n == 2) return 3; else return 2 * func(n - 1) + 1; }
程序运行:
实验任务四
程序源码:
#include<stdio.h> int func(int n, int m); int main() { int n, m; while (scanf_s("%d%d", &n, &m) != EOF) printf("n=%d,m=%d,ans=%d\n", n, m, func(n, m)); return 0; } int func(int n, int m) { if (n == 0 || n == m || m == 0) return 1; else if (n < m) return 0; else return func(n - 1, m) + func(n - 1, m - 1); }
程序运行:
实验任务五
task5.1
程序源码:
#include<stdio.h> double mypow(int x, int y); int main() { int x, y; double ans; while (scanf_s("%d%d", &x, &y) != EOF) { ans = mypow(x, y); printf("%d的%d次方:%g\n\n", x, y, ans); } return 0; } double mypow(int x, int y) { int t, i; if (y < 0) { t = -y;} else { t = y; } double sum = 1; for (i = 1; i <= t; i++) { sum *= x;} if (y > 0) { return sum; } else { 1.0 / sum; } }
程序运行:
task5.2
程序源码:
#include<stdio.h> double mypow(int x, int y); int main() { int x, y; double ans; while (scanf_s("%d%d", &x, &y) != EOF) { ans = mypow(x, y); printf("%d的%d次方:%g\n\n", x, y, ans); } return 0; } double mypow(int x, int y) { if (y == 0) { return 1; } if (y > 0) { return x * mypow(x, y - 1); } if (y < 0) { return 1.0 / x * mypow(x, y + 1); } }
程序运行:
实验任务六
程序源码:
#include<stdio.h> #include<stdlib.h> void hanoi(unsigned int n, char from, char temp, char to); void moveplate(unsigned int n, char from, char to); int tot; int main() { unsigned int n; while (scanf_s("%u", &n) != EOF) { tot = 0; hanoi(n, 'A', 'B', 'C'); printf("一共移动了%d次\n", tot);} system("pause"); return 0; } void hanoi(unsigned int n, char from, char temp, char to) { if (n == 1) { moveplate(n, from, to); } else { hanoi(n - 1, from, to, temp); moveplate(n, from, to); hanoi(n - 1, temp, from, to); } } void moveplate(unsigned int n, char from, char to) { tot ++; printf("%u:%c-->%c\n", n, from, to); }
程序运行:
实验任务七
程序源码:
#include<stdio.h> int is_prime(int x); int main() { int i, n; while (scanf_s("%d", &n) != EOF) { for (i = 2; i <= n; i++) { if (!is_prime(i) || !is_prime(n - i))continue; printf("%d = %d + %d\n", n, i, n - i); break; } } return 0; } int is_prime(int x) { int i; for (i = 2; i *i<= x; i++) { if (x % i == 0) { return 0;} else { return 1; } } }
程序运行:
实验任务八
程序源码:
#include<stdio.h> #include<math.h> long func(long s); int main() { long s, t; printf("Enter a number:"); while (scanf_s("%ld", &s) != EOF) { t = func(s); printf("new number is:%ld\n\n", t); printf("Enter a number:"); } return 0; } long func(long s) { long t = 0, t1 = 0; while (s > 0) { int tmp = s % 10; s /= 10; if (tmp % 2 == 1)t1 = t1 * 10 + tmp; } while (t1 > 0) { t = t * 10 + (t1 % 10); t1 /= 10; } return t; }
程序运行:
标签:return,int,long,char,实验,func,include From: https://www.cnblogs.com/HZJB1218/p/17273652.html