Test SRM Level Two: CountExpressions， Brute Force

题目来源：http://community.topcoder.com/stat?c=problem_statement&pm=8157

这道题目跟扑克牌算24的题目比较像，但要简单一些。点击查看 next_permutation 函数 的用法，使用这个函数减轻了不少工作量，而且算法很简洁明了。

代码如下：

 

#include <iostream>

#include <vector>

#include <algorithm>



using namespace std;



class CountExpressions

{

public:

	int calcExpressions(int x, int y, int val);

};



int CountExpressions::calcExpressions(int x, int y, int val)

{

	int i, j, k;

	int sum, count;

	vector <int> vnum;

	if (x > y) {

		swap(x, y);

	}

	/* 保证 x < y */

	vnum.push_back(x);

	vnum.push_back(x);

	vnum.push_back(y);

	vnum.push_back(y);

	

	sum = 0;

	count = 0;

	do {

		for (i = 0; i < 3; i++) {

		for (j = 0; j < 3; j++) {

		for (k = 0; k < 3; k++) {

			switch (i) {

			case 0:

				sum = vnum[0] + vnum[1];

				break;

			case 1:

				sum = vnum[0] - vnum[1];

				break;

			case 2:

				sum = vnum[0] * vnum[1];

				break;

			}



			switch (j) {

			case 0:

				sum = sum + vnum[2];

				break;

			case 1:

				sum = sum - vnum[2];

				break;

			case 2:

				sum = sum * vnum[2];

				break;

			}



			switch (k) {

			case 0:

				sum = sum + vnum[3];

				break;

			case 1:

				sum = sum - vnum[3];

				break;

			case 2:

				sum = sum * vnum[3];

				break;

			}



			if (sum == val) {

				++count;

			}

		}

		}

		}

	} while (next_permutation(vnum.begin(), vnum.end()));	/* 4个数字不同的排列组合 */



	return count;

}