24 Point game

标签：problem   space   括号   style   cout   exp   sed   led   else   

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。

输入

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

输出

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

样例输入

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

样例输出

Yes
No

来源

经典改编

上传者

张云聪

第一次由于用了vector数组覆盖的方法，太蠢了，大量的空间移动和数据拷贝，TLE

超时代码：

#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
/*
给一些数字，能否用利用+-*和/ 将这些数字组成一个特定的值，可以利用括号改变求值顺序
注意到给的数字个数很小 最多5个
每次将其中两个抽出来运算，然后将结果插入进去
*/
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;
}

 

24 Point game

标签：problem   space   括号   style   cout   exp   sed   led   else   

原文地址：http://www.cnblogs.com/joeylee97/p/6838045.html