Los solution to a problem Valley P1236 [24 point]

Have to say, personally I think that a lot of bigwigs want to program complex, so the amount of code is very long, but in fact, this problem does not have to be so complicated. . .

Could be considered a \ (DFS \) maintains a state \ (f (n) [a_1 , a_2 ...... a_n] \)

Next we enumerate violent match the two programs for each \ (A [I] \) , \ (A [J] \) larger value as long as the calculated value and a smaller number of two further analog arithmetic operation can be.

As mentioned in the comments section of the anti-reflection and anti besides actually having to consider, imagine every time we subtract the smaller value from the larger value, the larger value in addition to a small value, it is impossible for big or small decrease in addition to the small large.

For \ (A [I] \) , \ (a [j] \) match the two, may be the result of their operation into the \ (A [I] \) , and \ (a [j] \) can deposit \ (a [n] \) amount. It means simply scroll array of friends ~

This status is then compressed into \ (F (n--. 1) [A_1, A_2 ....... 1-n-A_ {}] \) , continue \ (DFS \) .

For the boundary processing, i.e., when the \ (n = 1 \) when the state becomes \ (F (. 1) [A_1] \) , \ (A_1 \) is the final result. If \ (a_1 = 24 \) illustrates the establishment of the maintenance process in each calculation \ (ok \) a.

#include<bits/stdc++.h>
using namespace std;
int num[10],f[20];
char ch[10];
bool Game(int n){
    if(n==1){//边界
        if(num[1]==24)return true;//判结果 
        else return false;   
    }
    for(int i=1;i<=n;i++){
        for(int j=i+1;j<=n;j++){//两两匹配,即可模拟所有可能
            int ti=num[i],tj=num[j];//保留原结果 
            int a=max(ti,tj);
            int b=min(ti,tj);//取较大较小值 
            int t=4-n; 
            num[j]=num[n];//压缩数组 
            //初始化
            f[t*3+1]=a;
            f[t*3+2]=b;
            f[t*3+3]=a+b; 
            ch[t+1]='+';//记录运算过程 
            num[i]=a+b;//做加法 
            if(Game(n-1))return true;//DFS 
            
            f[t*3+1]=a;
            f[t*3+2]=b;
            f[t*3+3]=a-b;   
            ch[t+1]='-';//记录运算过程 
            num[i]=a-b;//做减法
            if(Game(n-1))return true;//DFS 
           
            f[t*3+1]=a;
            f[t*3+2]=b;
            f[t*3+3]=a*b; 
            ch[t+1]='*';//记录运算过程 
            num[i]=a*b;//做乘法
            if(Game(n-1))return true;//DFS 
               
            if(b!=0&&a%b==0){//判断是否除的尽 
                if(a/b>=1){
                    f[t*3+1]=a;
                    f[t*3+2]=b;
                    f[t*3+3]=a/b; 
                    ch[t+1]='/';//记录运算过程 
                    num[i]=a/b;//做除法 
                    if(Game(n-1))return true;//DFS 
                }
            }
            
            num[i]=ti; 
            num[j]=tj;//答案错误还原之前结果 
        }
    } 
    return false;
}
int main(){
    for(int i=1;i<=4;i++){
        cin>>num[i];
    }
    if(Game(4)){
        cout<<f[1]<<ch[1]<<f[2]<<'='<<f[3]<<endl;
        cout<<f[4]<<ch[2]<<f[5]<<'='<<f[6]<<endl;
        cout<<f[7]<<ch[3]<<f[8]<<'='<<f[9]<<endl;//暴力输出 
    }else{
        cout<<"No answer!"<<endl;
    }
    return 0;
}