UVa 725 Division(暴力求解)

最新推荐文章于 2019-11-03 18:07:52 发布

YancyKahn

最新推荐文章于 2019-11-03 18:07:52 发布

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本文介绍UVa725题目“Division”的解决方案，该题要求找出所有符合条件的五位数对，使得其中一个五位数除以另一个等于指定整数N（2≤N≤79），且两个五位数恰好包含0到9的所有数字各一次。

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题目名称: Division
题目来源: UVa 725
题目链接:https://vjudge.net/problem/UVA-725
题目描述:
Write a program that finds and displays all pairs of 5-digit numbers that between them use the digits 0through 9 once each, such that the first number divided by the second is equal to an integer N, where2 ≤ N ≤ 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
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 <iostream>
#include <string.h>
//UVa 725
using namespace std;
struct Number{
    int arr[5];
    void cal(int num){
        int sum=1;
        for(int i = 0; i < 5; ++i){
            arr[i]=(num/sum)%10;
            sum*=10;
        }
    }
};
int main(){
    //abcde/fghij=n;
    int n,m,ans,x,count[10];
    bool flag=true;
    while(scanf("%d", &n) != EOF&&n != 0){
        x=0;
        if(flag){
            flag=false;
        }else{
            cout<<endl;
        }
        for(int i = 1234; i < 98765; i++){
        ans = i*n;
        Number nu[2];
        memset(count,0,sizeof(count));
        nu[0].cal(i);
        nu[1].cal(ans);
        if(ans > 98765) break;      
        for(int j = 0; j < 5; ++j){
            count[nu[0].arr[j]]++;
            count[nu[1].arr[j]]++;
        }
        for(m = 0; m <= 9; m++){
            //cout<<count[m]<<" ";
            if(count[m]>1){
                break;
            }
        }
        //cout<<endl;
        if(m == 10){
            printf("%d%d%d%d%d / %d%d%d%d%d = %d\n", nu[1].arr[4], nu[1].arr[3], nu[1].arr[2], nu[1].arr[1], nu[1].arr[0], nu[0].arr[4], nu[0].arr[3], nu[0].arr[2], nu[0].arr[1], nu[0].arr[0], n);
            x++;
        }else{
            continue;
        }   
    }
    if(x == 0){
        cout<<"There are no solutions for "<<n<<"."<<endl; 
    }
    }
    return 0;
}