题解：循环即可
题目地址：https://www.nowcoder.com/questionTerminal/2e6a898974064e72ba09d05a60349c9e
1 /** 2 * Copyright(c) 3 * All rights reserved. 4 * Author : Ycute 5 * Date : 2019-10-31-16.37.32 6 * Description : 模拟输出 7 */ 8 #include<iostream> 9 #include<cstdio> 10 #include<cmath> 11 #include<cstring> 12 #include<algorithm> 13 using namespace std; 14 15 //降序 16 int cmp0(int a,int b){ 17 return a>b; 18 } 19 //升序 20 bool cmp1(int a,int b){ 21 return a<b; 22 } 23 int main(){ 24 int in[4]; 25 int temp[4]; 26 int p; 27 scanf("%d",&p); 28 //scanf("%1d%1d%1d%1d",&in[0],&in[1],&in[2],&in[3]); 29 temp[0]=in[0]=p/1000; 30 temp[1]=in[1]=p/100%10; 31 temp[2]=in[2]=p%100/10; 32 temp[3]=in[3]=p%10; 33 while(1){ 34 if(in[0]==in[1]&&in[1]==in[2]&&in[2]==in[3]){ 35 printf("N - N = 0000 "); 36 break; 37 } 38 int t1[4]; 39 int t2[4]; 40 t1[0]=temp[0]; 41 t1[1]=temp[1]; 42 t1[2]=temp[2]; 43 t1[3]=temp[3]; 44 t2[0]=temp[0]; 45 t2[1]=temp[1]; 46 t2[2]=temp[2]; 47 t2[3]=temp[3]; 48 sort(t1,t1+4,cmp0); 49 sort(t2,t2+4,cmp1); 50 int num0=t1[0]*1000+t1[1]*100+t1[2]*10+t1[3]; 51 int num1=t2[0]*1000+t2[1]*100+t2[2]*10+t2[3]; 52 int num2=num0-num1; 53 temp[0]=num2/1000; 54 temp[1]=num2/100%10; 55 temp[2]=num2%100/10; 56 temp[3]=num2%10; 57 printf("%d%d%d%d - %d%d%d%d = %d%d%d%d ",t1[0],t1[1],t1[2],t1[3],t2[0],t2[1],t2[2],t2[3],temp[0],temp[1],temp[2],temp[3]); 58 if(temp[0]==6&&temp[1]==1&&temp[2]==7&&temp[3]==4) break ; 59 } 60 return 0; 61 }