(算法练习)——数字黑洞

《算法笔记》P152
用了好几个函数

代码:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;

void F(int a[],int b[]){
	sort(a,a+4);
	for(int i = 3;i >=0;i--){
		b[3-i] = a[i];
	}
}

int tonumber(int a[]){
	int sum = 0;
	for(int i = 0;i <4;i++){
		sum = sum + a[i]*pow(10,3-i);
	}
	return sum;
}

void toarry(int number,int a[]){
	for(int i = 0;i <4;i++){
		a[i] = number/(pow(10,3-i));
		number = number - a[i]* (pow(10,3-i)); 
	}
}
int main(){
	char str[10];
	int record1[10];
	int record2[10];
	scanf("%s",str);
	for(int i = 0;i <4;i++){
		record1[i] = str[i] - '0';
	}
	//record1是从小到大,record2是从大到小 
	F(record1,record2);
	int s = tonumber(record2) - tonumber(record1);
	while(s != 0){
		if(s == 6174){
			printf("%04d-%04d=%04d",tonumber(record2),tonumber(record1),s);
			break;
		}
		printf("%04d-%04d=%04d\n",tonumber(record2),tonumber(record1),s);
		memset(record1,0,sizeof(record1));
		toarry(s,record1);
		memset(record2,0,sizeof(record2));
		F(record1,record2);
		s = tonumber(record2) - tonumber(record1);
	}
	if(s == 0){
		printf("%04d-%04d=0000",tonumber(record2),tonumber(record1));
	}
	
	return 0;
	
	
}
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转载自blog.csdn.net/weixin_42377217/article/details/104039178