九度OJ 1036：Old Bill-优快云博客

本文介绍了一个程序，该程序解决了一个有趣的问题：根据中间三位可见数字和已知的火鸡总数，推测总价的首位和末位数字，同时计算每只火鸡的最大可能价格。此程序适用于总数量在1到99之间的火鸡，且总价由五位数字组成。

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题目描述：
Among grandfather’s papers a bill was found.
72 turkeys $_679_ The first and the last digits of the number that obviously represented the total price of those turkeys are replaced here by blanks (denoted _), for they are faded and are illegible. What are the two faded digits and what was the price of one turkey? We want to write a program that solves a general version of the above problem. N turkeys$ _XYZ_
The total number of turkeys, N, is between 1 and 99, including both. The total price originally consisted of five digits, but we can see only the three digits in the middle. We assume that the first digit is nonzero, that the price of one turkeys is an integer number of dollars, and that all the
turkeys cost the same price.
Given N, X, Y, and Z, write a program that guesses the two faded digits and the original price. In case that there is more than one candidate for the original price, the output should be the most expensive one. That is, the program is to report the two faded digits and the maximum price per turkey for the turkeys.
输入：
The first line of the input file contains an integer N (0

#include <cstdio>
using namespace std;
int main(){
    int n,x,y,z;
    while(scanf("%d",&n) != EOF){
        scanf("%d%d%d",&x,&y,&z);
        bool flag = true;
        for(int i = 9;i > 0;i--){
            for(int j = 9;j >= 0;j--){
                int tmp = 10000*i + 1000*x + 100*y + 10*z + j;
                if(tmp % n == 0){
                    printf("%d %d %d\n",i,j,tmp/n);
                    flag = false;
                    break;
                }
            }
            if(flag == false)break;
        }
        if(flag == true)printf("0\n");
    }
}