UVA725 Division【暴力枚举】

最新推荐文章于 2019-12-07 00:00:59 发布

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本文介绍了一个编程问题，即找出所有符合条件的五位数对 (abcde/fghij)，其中两个数使用0到9的所有不同数字，并且第一个数除以第二个数等于给定的整数N（2<=N<=79）。提供了输入输出样例及代码实现。

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暴力啊暴力啊~~~
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 <= N <= 79
. That is, abcde / fghig = 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

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<iomanip>
#include<cstring>
#include<vector>
#include<iterator>
using namespace std;
int main()
{
    int a,b,c,d,e,f,g,h,i,j,N,x,t,w,s,p[10],sum,n,ok=0;
    while(scanf("%d", &x)!=EOF&&x!=0)
    {
        if(ok==0)
            ok=1;
        else
            printf("\n");    //格式格式啊！！！ 为了满足有一行空格的格式。。。
        n=0;
        for(t=1234; t<=98765; t++){
            s=0;
            memset(p, 0, sizeof(p));
            j=t%10;
            i=(t/10)%10;
            h=(t/100)%10;
            g=(t/1000)%10;
            f=t/10000;        
            sum = x*t;
            if(sum>98765) break;
            e=sum%10;
            d=(sum/10)%10;
            c=(sum/100)%10;
            b=(sum/1000)%10;
            a=sum/10000;
            p[a]=1;
            p[b]=1;
            p[c]=1;
            p[d]=1;
            p[e]=1;
            p[f]=1;
            p[g]=1;
            p[h]=1;
            p[i]=1;
            p[j]=1;   //如果有重复，则s就不为10 即不输出
            for(w=0; w<10; w++){
                s+=p[w];
            }
            if(s==10){
                printf("%d%d%d%d%d / %d%d%d%d%d = %d\n", a,b,c,d,e,f,g,h,i,j,x);
                n++;
            }
            else
                continue;
        }
        if(n==0) printf("There are no solutions for %d.\n", x);
    }
    return 0;
}