PAT (Advanced Level) 1069. The Black Hole of Numbers (20)

简单题。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<queue>
#include<string>
#include<algorithm>
using namespace std;

int n;
int a[5];
int num1,num2;

int f(int num)
{
    memset(a,0,sizeof a);
    int tmp=num;
    int z=0;
    while(tmp) a[z++]=tmp%10,tmp=tmp/10;
    num1=0,num2=0;
    sort(a,a+4);
    for(int i=0;i<4;i++) num1=num1*10+a[i];
    for(int i=3;i>=0;i--) num2=num2*10+a[i];
    return num2-num1;
}

int main()
{
    scanf("%d",&n);
    if(f(n)==0) printf("%04d - %04d = %04d
",n,n,0);
    else
    {
        while(1)
        {
            int res=f(n);
            printf("%04d - %04d = %04d
",num2,num1,res);
            n=res;
            if(n==6174) break;
        }
    }
    return 0;
}
原文地址:https://www.cnblogs.com/zufezzt/p/5633415.html