A题：uva725暴力求解

Description:

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 . 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.

 

xxxxx / xxxxx =N

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

解析：通过枚举除数fghij可以算出abcde，然后将这几个数字字符排序一下查找是否所有数字都不相同。特别注意的是：当ab
cde+fghij超过十位时可以终止枚举。
代码如下：

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

int main() 
{
    int n, k= 0;
    char buf[99];
    while(scanf("%d", &n) == 1 && n) 
    { 
        
        int cnt = 0;
        if(k++) 
            printf("\n");
        for(int fghij = 1234; ; fghij++)
        {
            int abcde = fghij * n;   //通过枚举fghij的值确定abcde的值
            sprintf(buf, "%05d%05d", abcde, fghij);  //将这两个字符格式化
            if(strlen(buf) > 10)  //  循环终止条件：abcde+fghij大于十位
                break;
            sort(buf, buf+10);    //排序后，方便剔除有重复数字的buf。
            bool ok = true;
            for(int i = 0; i < 10; i++)
                if(buf[i] != '0' + i) 
                    ok = false;    
                if(ok) 
                {
                    cnt++;
                    printf("%05d / %05d = %d\n", abcde, fghij, n);  //输出格式要格外注意！！
                }
        }
        if(!cnt) 
            printf("There are no solutions for %d.\n", n);
    }
    return 0;
}

注意暴力解题时要认真分析，避免枚举过多超内存。