hdu 3699 10 福州 现场 J - A hard Aoshu Problem

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int per[5],temper[5];
char left[10],right[10],equ[10];
int index[5];
int del(char ch){
    return index[ch-‘A‘];
}
int getnum(char* str){
    int ans=0;
    for(int i=0;str[i];i++){
        ans*=10;
        ans+=per[del(str[i])];
    }
    return ans;
}
void cpy(int *a,int *b,int len){
    for(int i=0;i<len;i++)a[i]=b[i];
}
void pr(int lnum,int rnum,int equ,int op){
    /*printf("%d ",lnum);
    if(op==0)putchar(‘+‘);
    else if(op==1)putchar(‘-‘);
    else if(op==2)putchar(‘*‘);
    else if(op==3)putchar(‘/‘);
    printf(" %d = %d\n",rnum,equ);*/
}
int check(){
    int ans=0;
    int lnum=getnum(left),rnum=getnum(right),eqnum=getnum(equ);
    if(lnum+rnum==eqnum){pr(lnum,rnum,eqnum,0);ans++;}
    if(lnum-rnum==eqnum){pr(lnum,rnum,eqnum,1);ans++;}
    if(lnum*rnum==eqnum){pr(lnum,rnum,eqnum,2);ans++;}
    if(rnum!=0&&lnum/rnum==eqnum&&lnum%rnum==0){pr(lnum,rnum,eqnum,3);ans++;}
    return ans;
}
int dfs(int s,int ind,int rlen,int llen,int elen,int allen){
    int ans=0;
    per[s]=ind;
    if(s==allen-1){
        cpy(temper,per,allen);
        if(!((per[del(left[0])]==0&&llen>1)||(per[del(right[0])]==0&&rlen>1)||(per[del(equ[0])]==0&&elen>1)))
        ans+=check();
        while(next_permutation(per,per+allen)){
           // for(int i=0;i<allen;i++)printf("%d%c",per[i],i==allen-1?‘\n‘:‘ ‘);
            if((per[del(left[0])]==0&&llen>1)||(per[del(right[0])]==0&&rlen>1)||(per[del(equ[0])]==0&&elen>1))continue;
            ans+=check();
        }
        cpy(per,temper,allen);
    }
    else {
        for(int i=ind+1;i<10;i++)ans+=dfs(s+1,i,llen,rlen,elen,allen);
    }
    return ans;
}
int main(){
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%s%s%s",left,right,equ);
        int ans=0;
        for(int i=0;i<5;i++)per[i]=i;
        int llen=strlen(left),rlen=strlen(right),elen=strlen(equ);
        memset(index,-1,sizeof(index));
        int allen=0;
        for(int i=0;left[i];i++)if(index[left[i]-‘A‘]==-1)index[left[i]-‘A‘]=allen++;
        for(int i=0;right[i];i++)if(index[right[i]-‘A‘]==-1)index[right[i]-‘A‘]=allen++;
        for(int i=0;equ[i];i++)if(index[equ[i]-‘A‘]==-1)index[equ[i]-‘A‘]=allen++;
        for(int i=0;i<10;i++)ans+=dfs(0,i,llen,rlen,elen,allen);
        printf("%d\n",ans);
    }
    return 0;
}