UVA 725 - Division-优快云博客

本文链接：https://blog.youkuaiyun.com/fightforyourdream/article/details/8647480

本文介绍了一款程序设计挑战，该程序旨在找出所有符合条件的五位数对，这些数对使用0到9的所有数字各一次，并且第一个数除以第二个数的结果为整数N。文章详细说明了输入输出格式及示例。

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Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0 through 9 once each, such that the first number divided by the second is equal to an integer N, 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 integer N. 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 for N.". Separate the output for two different values of N by 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