24 Point game

24 Point game

時間限制： 3000 ms  |  內存限制：65535 KB

難度： 5

 

描述

There is a game which is called 24 Point game.

In this game , you will be given some numbers. Your task is to find an expression which have all the given numbers and the value of the expression should be 24 .The expression mustn't have any other operator except plus,minus,multiply,divide and the brackets. 

e.g. If the numbers you are given is "3 3 8 8", you can give "8/(3-8/3)" as an answer. All the numbers should be used and the bracktes can be nested. 

Your task in this problem is only to judge whether the given numbers can be used to find a expression whose value is the given number。

 

輸入

The input has multicases and each case contains one line
The first line of the input is an non-negative integer C(C<=100),which indicates the number of the cases.
Each line has some integers,the first integer M(0<=M<=5) is the total number of the given numbers to consist the expression,the second integers N(0<=N<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers is non-negative and less than 100

輸出

For each test-cases,output "Yes" if there is an expression which fit all the demands,otherwise output "No" instead.

樣例輸入

2
4 24 3 3 8 8
3 24 8 3 3

樣例輸出

Yes
No

來源

經典改編

上傳者

張雲聰

第一次因爲用了vector數組覆蓋的方法，太蠢了，大量的空間移動和數據拷貝，TLE

超時代碼：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN  103
#define LLL 1000000000
#define INF 1000000009
#define eps 0.00000001
/*
給一些數字，可否用利用+-*和/ 將這些數字組成一個特定的值，能夠利用括號改變求值順序
注意到給的數字個數很小 最多5個
每次將其中兩個抽出來運算，而後將結果插入進去
*/
double aim;
int n, T;
vector<double> v,tmp;
bool f;
void print(vector<double> &a)
{
    for (int i = 0; i < a.size(); i++)
    {
        if (i) printf(" ");
        printf("%lf", a[i]);
    }
    cout << endl;
}
void dfs(int x)
{
    if (x == 1)
    {
        //print(tmp);
        double che = tmp[0] - aim;
        if (tmp[0] - aim > -eps&&tmp[0] - aim < eps)
        {
            f = true;
            //cout << "sdaasddddddddddddddddddddddddddddddddddddddddddddddddddddd" << endl;
        }
        return;
    }
    //print(tmp);
    vector<double> h = tmp;
    double a, b, mul, add, sub1, sub2, div1, div2;
    for (int i = 0; i < x; i++)
    {
        for (int j = i + 1; j < x; j++)
        {
            tmp = h;
            a = tmp[i], b = tmp[j];
            mul = a*b, add = a + b;
            sub1 = a - b, sub2 = b - a;
            if (a != 0.0)
                div1 = b / a;
            else
                div1 = INF;
            if (b != 0.0)
                div2 = a / b;
            else
                div2 = INF;
            tmp.erase(tmp.begin() + i);
            tmp.erase(tmp.begin() + j - 1);
            vector<double> r = tmp;
            for (int i = 0; i <= tmp.size(); i++)
            {
                tmp.insert(tmp.begin() + i, add);
                dfs(tmp.size());
                tmp = r;
                if (f) return;
                tmp.insert(tmp.begin() + i, mul);
                dfs(tmp.size());
                tmp = r;
                if (f) return;
                tmp.insert(tmp.begin() + i, sub1);
                dfs(tmp.size());
                tmp = r;
                if (f) return;
                if (sub2-sub2>-eps&&sub1-sub2<eps)
                {
                    tmp.insert(tmp.begin() + i, add);
                    dfs(tmp.size());
                    tmp = r;
                }
                if (f) return;
                tmp.insert(tmp.begin() + i, div1);
                dfs(tmp.size());
                tmp = r;
                if (f) return;
                if (div1-div2<eps&&div1-div2>-eps)
                {
                    tmp.insert(tmp.begin() + i, div2);
                    dfs(tmp.size());
                    tmp = r;
                }
                if (f) return;
            }
        }
    }
}
int main()
{
    scanf("%d", &T);
    while (T--)
    {
        v.clear();
        scanf("%d%lf", &n, &aim);
        int t;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &t);
            v.push_back(t);
        }
        f = false;
        tmp = v;
        dfs(n);
        if (f)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}