POJ 1014--Dividing

来源：互联网发布：docker web mysql 编辑：程序博客网时间：2024/06/05 02:40

题目大意呢是说有若干个价值为1，2，3，4，5，6的大理石，问能否将这些大理石分成两份，让其总价值相等，大理石不可拆。挺简单的，直接深搜即可。代码如下：

Memory: 692K Time: 0MS Length：31Lines

#include<iostream>using namespace std;bool Recurring(const int sum, int remainder, int start, int* marbles, int* array){if (remainder == sum / 2)return true;for (int i = start; i >= 0; --i){if (array[i] < marbles[i] && remainder - i - 1 >= sum / 2){++array[i];if (Recurring(sum, remainder - i - 1, i, marbles, array))return true;}}return false;}int main(){int marbles[6];int count = 0;while (cin >> marbles[0] >> marbles[1] >> marbles[2] >> marbles[3] >> marbles[4] >> marbles[5]){++count;int sum = marbles[0] * 1 + marbles[1] * 2 + marbles[2] * 3 + marbles[3] * 4 + marbles[4] * 5 + marbles[5] * 6;if (sum == 0)break;int array[6] = { 0 };if (sum % 2 == 0 && Recurring(sum, sum, 5, marbles, array))cout << "Collection #" << count << ":" << endl << "Can be divided." << endl << endl;elsecout << "Collection #" << count << ":" << endl << "Can't be divided." << endl << endl;}return 0;}

0 0