1069 The Black Hole of Numbers

Posted 2021-09-23 CSU迦叶

注意两点：

1. 不足4位要补足，不仅仅是一开始要考虑，每次得到一个差值，都要考虑

2. 到0也会停下，不仅仅是一开始可能发生，也可能是过程中的某一个差值

另：

vector<int> 是可以作为函数的参数的。

AC代码

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<bits/stdc++.h>
#include<stdlib.h>
#include<time.h>
#include<vector>
#include<set>
#include<string>
#include<queue>
#include<map>

using namespace std;
typedef long long LL;

const int maxn = 50007;
const int MOD = 1000000007;
const int INF = 1000000000;//INF:下确界  
const LL SUP = (1LL<<63)-1;//SUP:上确界 
const double eps = 1e-5;

int getNum(vector<int> vi){
	int ans = 0;
	for(int i=vi.size()-1;i>=0;i--){
		ans = ans*10+vi[i];
	}
	return ans;	
}

int main(){
	string s;
	vector<int> vi; 
	cin>>s;
	
	while(s.length()!=4){//不足4位补零 
		s+="0";	
	}
	
	for(int i=0;i<s.length();i++){
		vi.push_back(s[i]-'0');
	}
	
	sort(vi.begin(),vi.end());

	int big = getNum(vi);
	reverse(vi.begin(),vi.end());
	int small = getNum(vi);
	int sub = big - small;
	
	while(sub!=6174&&sub!=0){
		printf("%04d - %04d = %04d\\n",big,small,sub);
		vi.clear();
		
		while(sub){
			vi.push_back(sub%10);
			sub /= 10;
		}
		while(vi.size()!=4){//不足4位补零
			vi.push_back(0);
		}	
		sort(vi.begin(),vi.end());	
		
		big = getNum(vi);
		reverse(vi.begin(),vi.end());
		small = getNum(vi);
		sub = big - small;
	}
	printf("%04d - %04d = %04d\\n",big,small,sub);
	
	return 0;
}