UVA 725 - Division-优快云博客

本文探讨了如何通过编程找到所有五位数与四位数的组合，使得五位数能被四位数整除，且每个数字在组合中各不相同。程序输入为整数N，输出符合条件的五位数与四位数的配对，或当无解时提示无解。

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Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits0through9once each, such that the first number divided by the second is equal to an integerN, where $2\le N \le 79$ . That is,

abcde / fghij =N

where each letter represents a different digit. The first digit of one of the numerals is allowed to be zero.

Input

Each line of the input file consists of a valid integerN. An input of zero is to terminate the program.

Output

Your program have to display ALL qualifying pairs of numerals, sorted by increasing numerator (and, of course, denominator).

Your output should be in the following general form:

xxxxx / xxxxx =N

In case there are no pairs of numerals satisfying the condition, you must write ``There are no solutions forN.". Separate the output for two different values ofNby a blank line.

Sample Input

61
62
0

Sample Output

There are no solutions for 61.

79546 / 01283 = 62
94736 / 01528 = 62

#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>

using namespace std;

#define MAXN 1000



bool check(int a, int b){
	int repeat[11] = {0};

	if(a < 10000){
		repeat[0]++;
	}
	if(b < 10000){
		repeat[0]++;
	}

	if(repeat[0] > 1){
		return false;
	}

	while(a != 0){
		int remain = a % 10;
		if(repeat[remain] != 0){
			return false;
		}
		repeat[remain]++;
		a /= 10;
	}

	while(b != 0){
		int remain = b % 10;
		if(repeat[remain] != 0){
			return false;
		}
		repeat[remain]++;
		b /= 10;
	}

	return true;
}

void play(int n){

	int fghij = 1234;
	int abcde = n * fghij;

	bool found = false;
	for(; abcde<100000; ++fghij){
		abcde = n * fghij;
		if(check(abcde, fghij)){
			found = true;
			//自动补零
			printf("%05d / %05d = %d\n", abcde, fghij, n);
		}
	}

	if(!found){
		printf("There are no solutions for %d.\n", n);
	}
}


int main(){

#ifndef ONLINE_JUDGE
	freopen("725.in", "r", stdin);
	freopen("725.out", "w", stdout); 
#endif

	int n;

	// 有用的技巧来避免结尾多输出一空行
	bool blank = false;

	while(scanf("%d",&n)==1 && n){
		if(blank){
			printf("\n");
		}
		play(n);
		blank = true;
	}


}


#endif