HDU 1247 速算24点

#include <iostream>
#include <cstdio>
#include <cstring>
#define CL(x, y) memset(x, y, sizeof(x))
using namespace std;
const int INF = 1 << 30;
int num[4], res[4], used[4];
int flag;
int getNum(char ch);
int calculate(int x,int y,int i);  //4步运算
int checkNum();        //判断24
void DFS(int k);    //全排列
int main()
{
    char ch;
    int i, j, k;
    while(cin>>ch)
    {
        for(i = 0; i < 4; i++)//输入有点讲究，十分重要
        {
            if(ch=='1')
            {
                num[i] = 10;
                cin >> ch;//输入为10的时候
            }
            else
            {
                num[i] = getNum(ch);//将当前输入的字符转化为数字
            }
            if(i<3)
                cin>>ch;
        }
        flag = 0;
        CL(used, 0);
        DFS(0);
        if(flag) printf("Yes
");
        else printf("No
");
    }
    return 0;
}
int getNum(char ch)
{
    switch(ch)
    {
    case 'A':
        return 1;
    case 'J':
        return 11;
    case 'Q':
        return 12;
    case 'K':
        return 13;
    default:
        return ch-'0';
    }
}
int calculate(int x,int y,int i)  //4步运算
{
    switch(i)
    {
    case 0 :
        return x + y;
    case 1 :
        return x - y;
    case 2 :
        return x * y;
    case 3 :
        if(y != 0 && !(x % y))
            return x / y;
        else
            return INF;
    }
}
void DFS(int k)    //全排列24
{
    int i;
    if(k == 4)
    {
        if(checkNum()) flag = 1;
        return;
    }
    for(i = 0; i < 4; i ++)
    {
        if(!used[i])
        {
            res[k] = num[i];
            used[i] = 1;
            DFS(k + 1);
            if(flag) return;
            used[i] = 0;
        }
    }
}
int checkNum()        //判断24
{
    int i,j,k;
    int t1,t2,t3;
    for(i = 0; i < 4; i ++)       //(a @ b) @ (c @ d)
    {
        t1 = calculate(res[0],res[1],i);
        if(t1 == INF) continue;
        for(j = 0; j < 4; j ++)
        {
            t2 = calculate(res[2],res[3],j);
            if(t2 == INF) continue;
            for(k = 0; k < 4; k ++)
            {
                t3 = calculate(t1,t2,k);
                if(t3 == 24) return 1;
            }
        }
    }
    for(i = 0; i < 4; i ++)   //((a @ b) @ c) @ d
    {
        t1 = calculate(res[0],res[1],i);
        if(t1 == INF) continue;
        for(j = 0; j < 4; j ++)
        {
            t2 = calculate(t1,res[2],j);
            if(t2 == INF) continue;
            for(k = 0; k < 4; k ++)
            {
                t3 = calculate(t2,res[3],k);
                if(t3 == 24) return 1;
            }
        }
    }
    for(i = 0; i < 4; i ++)   //(a @ (b @ c)) @ d
    {
        t1 = calculate(res[1],res[2],i);
        if(t1 == INF) continue;
        for(j = 0; j < 4; j ++)
        {
            t2 = calculate(res[0],t1,j);
            if(t2 == INF) continue;
            for(k = 0; k < 4; k ++)
            {
                t3 = calculate(t2,res[3],k);
                if(t3 == 24) return 1;
            }
        }
    }
    for(i = 0; i < 4; i ++)   //a @ (b @ (c @ d))
    {
        t1 = calculate(res[2],res[3],i);
        if(t1 == INF) continue;
        for(j = 0; j < 4; j ++)
        {
            t2 = calculate(res[1],t1,j);
            if(t2 == INF) continue;
            for(k = 0; k < 4; k ++)
            {
                t3 = calculate(res[0],t2,k);
                if(t3 == 24) return 1;
            }
        }
    }
    for(i = 0; i < 4; i ++)   //a @ ((b @ c) @ d)
    {
        t1 = calculate(res[1],res[2],i);
        if(t1 == INF) continue;
        for(j = 0; j < 4; j ++)
        {
            t2 = calculate(t1,res[3],j);
            if(t2 == INF) continue;
            for(k = 0; k < 4; k ++)
            {
                t3 = calculate(res[0],t2,k);
                if(t3 == 24) return 1;
            }
        }
    }
    return 0;    //无解
}

24点问题：

先不考虑括号，4个数是全排列，则有4！；

3个运算符，每个运算符有4次选择机会，一共有4^3；

接下来考虑括号的问题，则有5种方案：(A(B(CD)))，(A((BC)D))，((AB)(CD))，((A(BC))D)，(((AB)C)D)

综上，一共有：4！* 4^3 * 5 = 1536种表达式；