#include<iostream>
#include<algorithm>
using namespace std;
int n;
int cz(int n,bool m){
	int k;
	int a[]={n/1000,n/100%10,n/10%10,n%10};
	sort(a,a+4);
	if(!m)k=a[0]*1000+a[1]*100+a[2]*10+a[3];//如果m为false,返回最小数
	else k=a[3]*1000+a[2]*100+a[1]*10+a[0];//否则输出最大数
	return k;
}
int main(){
	cin>>n;
	while(n!=6174){
		int a=cz(n,1),b=cz(n,0);
		cout<<a<<"-"<<b<<"="<<a-b<<endl;
		n=a-b;
	}
	return 0;
}

#include <iostream> #include <algorithm> #include <cstring> using namespace std;

int n,a[6],b[6];

bool cmp(int x,int y) { return x > y; }

int main() { cin >> n; while(n != 6174) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); int num = n,sum = 0; n = 0; for(int i = 1;num != 0;i++) { a[i] = num%10; b[i] = num%10; num /= 10; } sort(a+1,a+5,cmp); sort(b+1,b+5); for (int i = 1;i <= 4;i++) { n = n10+a[i]; sum = sum10+b[i]; } cout << n << "-" << sum << "=" << n-sum << endl; n = n-sum; } return 0; }

#include <iostream> #include <cstdio> #include <cmath> #include <algorithm> #include <stack> #include <queue> #include <map> #include <cstring> #include <iomanip> using namespace std; int n; int cz(int n,bool m){ int k; int a[]={n/1000,n/100%10,n/10%10,n%10}; sort(a,a+4); if(!m)k=a[0]*1000+a[1]*100+a[2]*10+a[3]; else k=a[3]*1000+a[2]*100+a[1]*10+a[0]; return k; } int main(){ cin>>n; while(n!=6174){ int a=cz(n,1),b=cz(n,0); cout<<a<<"-"<<b<<"="<<a-b<<endl; n=a-b; }

return 0;}

#include<bits/stdc++.h>
using namespace std;
int n;
int cz(int n,bool m){
	int k;
	int a[]={n/1000,n/100%10,n/10%10,n%10};
	sort(a,a+4);
	if(!m)k=a[0]*1000+a[1]*100+a[2]*10+a[3];
	else k=a[3]*1000+a[2]*100+a[1]*10+a[0];
	return k;
}
int main(){
	cin>>n;
	while(n!=6174){
		int a=cz(n,1),b=cz(n,0);
		cout<<a<<"-"<<b<<"="<<a-b<<endl;
		n=a-b;
	}
	return 0;
}

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int main(){
    int n;
    cin >> n;
    while(true){
        char a[5], a2[5];
        sprintf(a, "%d", n); //将数字转换为字符数组 
        sprintf(a2, "%d", n);
        sort(a, a+strlen(a), less<int>()); //重组最大数
        sort(a2, a2+strlen(a2), greater<int>());//重组最小数
        int m = atoi(a); //将字符串转换为int 
        n = atoi(a2);
        int ans = n - m;//相减结果 
        printf("%d-%d=%d\n", n, m, ans);
        n = ans;
        if(ans == 6174)
            break; //遇到6174时结束循环 
    }
    return 0;
}

/*****************************************
备注：
******************************************/
#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int main()
{
	int n ;
	cin >> n;
	while(n != 6174)
	{
		int a[10];
		int m = n;
		memset(a,0,sizeof(a));
		while(m)
		{
			a[m%10]++;
			m/=10;
		}
		int ans =0;
		for(int i = 9 ; i >=0 ; i--)
		{
			for(int j= 0 ;j < a[i] ; j++)
			{
				ans = ans *10 + i;
			}
		}
		int sum = 0;
		for(int i = 0 ; i < 10 ; i++)
		{
			for(int j= 0 ;j < a[i] ; j++)
			{
				sum = sum *10 + i;
			}
		}
		cout <<ans << "-"<< sum << "=" << ans-sum<<endl;
		n = ans-sum;
	}
	return 0;
}