暴力枚举 + 24点 --- hnu : Cracking the Safe

Cracking the Safe

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 46, Accepted users: 12

Problem 12886 : No special judgement

Problem description

Secret agent Roger is trying to crack a safe containing evil Syrian chemical weapons. In order to crack the safe, Roger needs to insert a key into the safe. The key consists of four digits. Roger has received a list of possible keys from his informants that he needs to try out. Trying out the whole list will take too long, so Roger needs to find a way to reduce the list. A valid key satisfies a certain condition, which we call the 24 condition. Four digits that satisfy the 24 condition can be manipulated using addition, subtraction, multiplication, division and parentheses, in such a way, that the end result equals 24. For example, the key (4; 7; 8; 8) satisfies the 24 condition, because (7-8/8)×4 = 24. The key (1; 1; 2; 4) does not satisfy the 24 condition, nor does (1; 1; 1; 1). These keys cannot possibly be the valid key and do not need to be tried. Write a program that takes the list of possible keys and outputs for each key whether it satisfies the 24 condition or not.

Input

On the first line one positive number: the number of test cases, at most 100. After that per test case:  one line with four space-separated integers a; b; c; d (1<=a; b; c; d<=9): a possible key.

Output

Per test case:  one line with either “YES” or “NO”, indicating whether the key satisfies the 24 condition or not.

Sample Input

4 4 7 8 8 1 1 2 4 1 1 1 1 1 3 4 6

Sample Output

YES NO NO YES

Problem Source

BAPC preliminary 2013

---

 

Mean:

 

给你4个数，你需要判断这4个数是否能够通过"+"、"-"、"*"、"/"四种得到24。

 

 

analyse:

数据这么小，直接暴力枚举。

先枚举四个数的位置，再枚举三个运算符，最后枚举运算顺序。

解法二：递归求解。

 

Time complexity：4*4*4*4*4*4*4*5=81920，不超过81920次

 

Source code：

 

//Memory   Time
// 1143K   27MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
bool mark;
double num[5];
double aa[5];
double calc(double a,double b,int flag)
{
    switch(flag)
    {
        case 1:return (a+b);
        case 2:return (a-b);
        case 3:return (a*b);
        case 4:return (a/b);
    }
}
void algebra()
{
    double tmp1,tmp2,tmp3;
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            for(int k=1;k<=4;k++)
            {
                //  运算顺序1
                tmp1=calc(aa[1],aa[2],i);
                tmp2=calc(tmp1,aa[3],j);
                tmp3=calc(tmp2,aa[4],k);
                if(fabs(tmp3-24)<1e-6)
                {
                   mark=1;
                    return ;
                }
                //  运算顺序2
                tmp1=calc(aa[1],aa[2],i);
                tmp2=calc(aa[3],aa[4],k);
                tmp3=calc(tmp1,tmp2,j);
                if(fabs(tmp3-24)<1e-6)
                {
                   mark=1;
                    return ;
                }
                //  运算顺序3
                tmp1=calc(aa[2],aa[3],j);
                tmp2=calc(aa[1],tmp1,i);
                tmp3=calc(tmp2,aa[4],k);
                if(fabs(tmp3-24)<1e-6)
                {
                   mark=1;
                    return ;
                }
                //  运算顺序4
                tmp1=calc(aa[2],aa[3],j);
                tmp2=calc(tmp1,aa[4],k);
                tmp3=calc(tmp2,aa[1],i);
                if(fabs(tmp3-24)<1e-6)
                {
                   mark=1;
                    return ;
                }
                //  运算顺序5
                tmp1=calc(aa[3],aa[4],k);
                tmp2=calc(aa[2],tmp1,j);
                tmp3=calc(aa[1],tmp2,i);
                if(fabs(tmp3-24)<1e-6)
                {
                   mark=1;
                    return ;
                }
            }
        }
    }
}

void arrange()
{
    for(int i=1;i<=4;i++)
    {
        for(int j=1;j<=4;j++)
        {
            if(j==i)continue;
            for(int k=1;k<=4;k++)
            {
                if(k==i||k==j)continue;
                for(int l=1;l<=4;l++)
                {
                    if(l==i||l==j||l==k)continue;
                    aa[1]=num[i],aa[2]=num[j],aa[3]=num[k],aa[4]=num[l];
                    algebra();
                }
            }
        }
    }
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        mark=false;
        for(int i=1;i<=4;i++)
            cin>>num[i];
        arrange();
        if(mark)
            puts("YES");
        else
            puts("NO");
    }
    return 0;
}

 

　　递归：

//Memory   Time
// 724K      0MS
// by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAX 1100
#define LL long long
using namespace std;
double num[4];
bool solve ( int n ) {
    if ( n == 1 ) {
        if ( fabs ( num[0] - 24 ) < 1E-6 )
            return true;
        else
            return false;
    }

    for ( int i = 0; i < n; i++ ) 
    {
        for ( int j = i + 1; j < n; j++ ) 
        {
                    double a, b;
                    a = num[i];
                    b = num[j];
                    num[j] = num[n - 1];
                    
                    num[i] = a + b;
                    if ( solve ( n - 1 ) )
                        return true;
                        
                    num[i] = a - b;
                    if ( solve ( n - 1 ) )
                        return true;

                    num[i] = b - a;
                    if ( solve ( n - 1 ) )
                        return true;

                    num[i] = a * b;
                    if ( solve ( n - 1 ) )
                        return true;
                        
                        
                    if ( b != 0 ) {
                        num[i] = a / b;

                        if ( solve ( n - 1 ) )
                            return true;
                    }
                    if ( a != 0 ) {
                        num[i] = b / a;
                        if ( solve ( n - 1 ) )
                            return true;
                    }
                    num[i] = a;
                    num[j] = b;
        }
    }
    return false;
}
int main() {
    int x;
    int T;
    cin >> T;
    while ( T-- ) {
        for ( int i = 0; i < 4; i++ ) {
            cin >> x;
            num[i] = x;
        }

        if ( solve ( 4 ) ) {
            cout << "YES" << endl;
        } else {
            cout << "NO" << endl;
        }
    }

    return 0;
}