hdu 1427 速算24点

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1427 

速算24点

Description

速算24点相信绝大多数人都玩过。就是随机给你四张牌，包括A(1),2,3,4,5,6,7,8,9,10,J(11),Q(12),K(13)。要求只用'+','-','*','/'运算符以及括号改变运算顺序，使得最终运算结果为24(每个数必须且仅能用一次)。游戏很简单，但遇到无解的情况往往让人很郁闷。你的任务就是针对每一组随机产生的四张牌，判断是否有解。我们另外规定，整个计算过程中都不能出现小数。

Input

每组输入数据占一行，给定四张牌。

Output

每一组输入数据对应一行输出。如果有解则输出"Yes"，无解则输出"No"。

Sample Input

A 2 3 6
3 3 8 8

Sample Output

Yes
No

参照<<编程之美>>上的写法，时间卡的挺紧的，用cin tle scanf过了/(ㄒoㄒ)/~~

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
using std::map;
using std::fabs;
using std::pair;
using std::swap;
using std::string;
using std::next_permutation;
#define pb(e) push_back(e)
#define sz(c) (int)(c).size()
#define mp(a, b) make_pair(a, b)
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr, val) memset(arr, val, sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for(int i = 0; i < (int)n; i++)
#define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
const int N = 20;
const double eps = 1E-7;
const int INF = 0x3f3f3f3f;
typedef unsigned long long ull;
int arr[4];
char ret[4][10];
map<string, int> A;
bool dfs(int n) {
	if (1 == n) {
		return arr[0] == 24;
	}
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			int a = arr[i], b = arr[j];
			arr[j] = arr[n - 1];
			arr[i] = a + b;
			if (dfs(n - 1)) return true;
			arr[i] = a - b;
			if (dfs(n - 1)) return true;
			arr[i] = b - a;
			if (dfs(n - 1)) return true;
			arr[i] = a * b;
			if (dfs(n - 1)) return true;
			if (b != 0 && 0 == a % b) {
				arr[i] = a / b;
				if (dfs(n - 1)) return true;
			}
			if (a != 0 && 0 == b % a) {
				arr[i] = b / a;
				if (dfs(n - 1)) return true;
			}
			arr[i] = a;
			arr[j] = b;
		}
	}
	return false;
}
void init() {
	A["A"] = 1;
	A["2"] = 2;
	A["3"] = 3;
	A["4"] = 4;
	A["5"] = 5;
	A["6"] = 6;
	A["7"] = 7;
	A["8"] = 8;
	A["9"] = 9;
	A["10"] = 10;
	A["J"] = 11;
	A["Q"] = 12;
	A["K"] = 13;
}
bool solve() {
	for (int i = 0; i < 4; i++) {
		arr[i] = A[ret[i]];
	}
	do {
		if (dfs(4)) return true;
	} while (next_permutation(arr, arr + 4));
	return false;
}
int main() {
#ifdef LOCAL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w+", stdout);
#endif
	init();
	while (~scanf("%s %s %s %s", ret[0], ret[1], ret[2], ret[3])) {
		puts(solve() ? "Yes" : "No");
	}
	return 0;
}

转载于:https://www.cnblogs.com/GadyPu/p/4895433.html