除法（暴力）

除法

Description

 

Division

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.

 

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

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


题意：
输入正整数n，按从小到大的顺序输出所有形如abcde/fghij=n的表达式，其中a-j恰好为数字0-9的一个排列（可以有前导0）， 。

分析：
1.枚举fghij就可以算出abcde,只需遍历所有的分子即可
2.判断是否所有数字都不相同，数字最大为98765，最小为1234.
3.当abcde和fghij加起来超过10位时可以终止枚举
4.暴力求解

代码：

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

bool tese(int i,int j)    //用数组a存放i，j的各位数字
{
    int a[10]={0};        //初始化数组a，使得各位数字为0，使fghij<10000 时f位置为0
    int t=0;
    while(i)              //取i中各位数字存放在数组a中
    {
        a[t++]=i%10;
        i=i/10;
    }
    while(j)              //取j中各位数字存放在数组a中
    {
        a[t++]=j%10;
        j=j/10;
    }
    //判断a～j是否恰好为数字的0～9的一个排列
    for(int m=0;m<=10;++m)
        for(int n=m+1;n<10;++n)
            if(a[n]==a[m])
                return 0;
            return 1;
}
int main()
{
    int n,k,s=0;
    while(scanf("%d",&n)&&n>=2&&n<=79)
    {
        if(s++)          //输出一行空行
            cout<<endl;
        int count=0;
        for(k=1234;k<=98765;k++)
        {
            int l=k*n;
            if(l<100000)
            {
                if(tese(l,k))
                {
                    printf("%05d / %05d = %d\n",l,k,n);
                   count=1;
                }
            }
            else if(l>100000&&count!=1)
            {
                printf("There are no solutions for %d.\n",n);
                break;
            }            
        }
    }
    return 0;
}

心得：

以前只是听过用暴力方法解题的，不知道暴力是什么，现在真正用到暴力，发现暴力确实很简单。继续暴力吧。