KY95 Old Bill

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.
输入描述：
The first line of the input file contains an integer N (0<N<100), which represents the number of turkeys. In the following line, there are the three decimal digits X, Y, and Z., separated by a space, of the original price $XYZ.
输出描述：
For each case, output the two faded digits and the maximum price per turkey for the turkeys.
示例1

题目大意：
N只火鸡的价格为$_XYZ_，火鸡的总数N在1到99之间。价格由五位数组成，两边的数字由于褪色而看不清，所以只能看到中间的三位数。假设第一个数字非零，每只火鸡的价格是整数，并且所有火鸡的价格相同。给定N，X，Y和Z，编写一个程序来猜测两边褪色的数字和火鸡的原始价格。如果多个价格符合题意，那么输出最昂贵的那个。
知识点： 暴力求解枚举
AC代码：

#include<bits/stdc++.h>using namespace std;int main()
{int n;int x, y, z;int sum;int a, b;while(cin >> n) // 多样例输入注意用while循环输入{cin >> x >> y >> z;int flag = 0;for(int i = 9; i > 0; i--) // 做减循环，得到最贵的就可以跳出{for(int j = 9; j >= 0; j--){sum = i * 10000 + x * 1000 + y * 100 + z * 10 + j;if(sum % n == 0){cout << i << ' ' << j << ' ' << sum / n << endl;flag = 1;break; //跳出内层循环}}if(flag){break; // 跳出外层循环}}if(flag == 0) // 注意：找不到结果时，输出0{cout << 0 << endl;}}return 0;
}

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