实验三

一

运行代码为

#include<stdio.h> #include<stdlib.h> #include<time.h> #include<windows.h> #define N 80 void printText(int line, int col, char text[]); void printSpaces(int n); void printBlankLines(int n); int main() {     int line,col,i;     char text[N] = Hi,May~~;          srand(time(0));      for (i = 0; i <= 10; i++)     {         line = rand() % 25;         col = rand() % 80;         printText(line, col, text);         Sleep(1000);     }      return 0; } void printBlankLines(int n) {     int i;     for (i = 1; i <= n; i++)         printf(\n); } void printSpaces(int n) {     int i;     for (i = 1; i <= n; i++)         printf( ); } void printText(int line, int col, char text[]) {     printBlankLines(line - 1);     printSpaces(col - 1);     printf(%s, text); }

功能

随机生成数0-24和0-80的数赋值到line和col，是打印“hi may”时前打印col-1个空格，且每个“hi，may”间隔line-1行、且每生成一个”Hi may“后做1000ms的停留。

二

1.运行原源代码

#include<stdio.h> long long fac(int n); int main() {     int i, n;      printf(enter n:);     scanf(%d, &n);      for (i = 1; i <= n; i++)         printf(%d! = %11d\n, i, fac(i));      return 0; } long long fac(int n) {     static long long p = 1;      p = p * n;      return p; }

添加“printf(p = %lld\n, p);”在17与19行间，运行结果为

2.运行代码为

 

#include<stdio.h> long func(int ,int ); int main() {     int k = 4, m = 1, p1, p2;      p1 = func(k, m);     p2 = func(k, m);     printf(%d,%d, p1, p2);     return 0; } int func(int a, int b) {     static int m = 0, i = 2;     i += m + 1;     m = i + a + b;      return m; }

运行结果为

static变量将保存上次运行的结果进入下一次运算

三

#include<stdio.h> long long fun(int n); int main() {     int n;     long long f;     while (scanf(%d, &n) != EOF)     {         f = fun(n);         printf(n=%d,f=%lld\n, n, f);     }     return 0; } long long fun(int n) {     long long a;     if (n == 0)         return 0;     else if (n == 1)         return 1;     if (n != 0)         return 2 * fun(n - 1) + 1; }

运行结果为

四

运行代码为

 

#include<stdio.h> void han (int n); long long fun(int n); int main() {     int n,p;     char A='A',B='B',C='C';     while (scanf(%d, &n) != EOF)     {         han(n,A,C,B);         p = fun(n);         printf(一共移动了%d次\n, p);     }     return 0; } void han(int n,char x,char y,char z) {     static int p=0;     if (n == 1)     {         printf(第%d个盘子：%c-->%c\n, n, x, y);     }     if (n != 1)     {         han(n-1,x,z,y);         printf(第%d个盘子：%c-->%c\n,n,x,y);         han(n-1,z,y,x);     } } long long fun(int n) {     long long a;     if (n == 1)         return 1;     if (n != 0)         return 2 * fun(n - 1) + 1; }

运行结果为

 五

运行代码为

#include<stdio.h> int is_prime(int x); int main() {     int j = 1,p,q,k;     for (j = 0; j <= 20; j=j+2)     {         for (p = 1;p < j; p++)         {             k = is_prime(p);             if (k == 1)             {                 q = j - p;                 k = is_prime(q);                 if (k == 1)                 {                     printf(%d=%d+%d\n, j, p, q);                     break;                 }             }                      }     }     return 0; } int is_prime(int x) {     int i;     for (i = 2; i < x; i++)         if (x % i == 0)break;     if (i == x)         return 1;     else         return 0; }

运行结果为

六

运行代码为

#include<stdio.h> long fun(long s); int main() {     long s, t;     printf(enter a number:);     while (scanf(%ld, &s) != EOF)     {         t = fun(s);         printf(new number is: %1d\n\n, t);         printf(enter a number:);     }     return 0; } long fun(long s) {     int m=1,k,s0=0;     while (s != 0)     {         k = s % 10;         if (k % 2 == 1)         {             s0 = s0 + k * m;             m = 10 * m;         }         s = s / 10;     }     return s0; }

运行结果为