UVa 10344 23 out of 5 (全排列枚举&回溯)

最新推荐文章于 2019-10-23 20:45:52 发布

最新推荐文章于 2019-10-23 20:45:52 发布 · 134 阅读

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本文提供了一种高效的算法解决方案，用于确定五个给定整数中是否存在组合，通过加减乘操作得到23。通过排序和递归搜索，实现快速判断可能性。

10344 - 23 out of 5

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1285

Your task is to write a program that can decide whether you can find an arithmetic expression consisting of five given numbers(1<=i<=5) that will yield the value 23.
For this problem we will only consider arithmetic expressions of the following from:

where : {1,2,3,4,5} -> {1,2,3,4,5} is a bijective function

and {+,-,*} (1<=i<=4)

Input

The Input consists of 5-Tupels of positive Integers, each between 1 and 50.
Input is terminated by a line containing five zero's. This line should not be processed.

Output

For each 5-Tupel print "Possible" (without quotes) if their exists an arithmetic expression (as described above) that yields 23. Otherwise print "Impossible".

Sample Input

1 1 1 1 1

1 2 3 4 5

2 3 5 7 11

0 0 0 0 0

Sample Output

Impossible

Possible

Possible

水题。

完整代码：

/*0.279s*/

#include<cstdio>
#include<algorithm>
using namespace std;

int a[5];
bool flag;

void dfs(int cur, int ans)
{
	if (cur == 5)
	{
		if (ans == 23) flag = true;
		return;
	}
	if (!flag) dfs(cur + 1, ans + a[cur]);
	if (!flag) dfs(cur + 1, ans - a[cur]);
	if (!flag) dfs(cur + 1, ans * a[cur]);
}

int main()
{
	while (scanf("%d%d%d%d%d", &a[0], &a[1], &a[2], &a[3], &a[4]), a[0])
	{
		flag = false;
		sort(a, a + 5);
		do
		{
			dfs(1, a[0]);
			if (flag) break;
		}
		while (next_permutation(a, a + 5));
		puts(flag ? "Possible" : "Impossible");
	}
	return 0;
}