递归

来源：互联网发布：站内搜索数据库编辑：程序博客网时间：2024/06/14 17:56

本程序使用递归，在输入任意的数字n（1~12），都可以计算出 2 + 22 + 222 + ...... + 222...222的结果。
其中测试用例有：
n = 1 , result = 2;
n = 2 , result = 24;
n = 3 , result = 246;
n = 4 , result = 2468;
n = 5 , result = 24690;
n = 6 , result = 246912;
......
n = 12 , result = 2100444372;

但当 n = 13 , result = -470392734;超出了计算机的运算范围。

#include<stdio.h>#include<stdlib.h>int fun(int n);int result(int n);int main(void){     int flag = 1;  int n;  int sumresult;  while(flag)  { system("cls"); printf("please input the number: "); scanf("%d",&n);     sumresult = result(n); printf("the result is  %d\n",sumresult); printf("Whether to continue testing, yes 1   no 0  :  ") ;      scanf("%d" , &flag);     }   system("PAUSE");  return 0;}int result(int n){  int sum = 0;  for(int i = 1; i <= n ; i++)  {    sum = sum + fun(i);  }  return sum ;   // printf("the result is  %d\n",sum);}int fun(int n){int temp = 0;    if(n == 1){   temp = 2;   return temp;}    else{  temp = fun(n-1)*10+2;  return temp;}}

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