杭电3699A hard Aoshu Problem DFS 搜索

A hard Aoshu Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 1521    Accepted Submission(s): 800

Problem Description

Math Olympiad is called “Aoshu” in China. Aoshu is very popular in elementary schools. Nowadays, Aoshu is getting more and more difficult. Here is a classic Aoshu problem:


ABBDE __ ABCCC = BDBDE

In the equation above, a letter stands for a digit(0 – 9), and different letters stands for different digits. You can fill the blank with ‘+’, ‘-‘ , ‘×’ or ‘÷’.

How to make the equation right? Here is a solution:


12245 + 12000 = 24245

In that solution, A = 1, B = 2, C = 0, D = 4, E = 5, and ‘+’ is filled in the blank.

When I was a kid, finding a solution is OK. But now, my daughter’s teacher tells her to find all solutions. That’s terrible. I doubt whether her teacher really knows how many solutions are there. So please write a program for me to solve this kind of problems.

 

Input

The first line of the input is an integer T( T <= 20) indicating the number of test cases.

Each test case is a line which is in the format below:


s1 s2 s3

s1, s2 and s3 are all strings which are made up of capital letters. Those capital letters only include ‘A’,’B’,’C’,’D’ and ‘E’, so forget about ‘F’ to ‘Z’. The length of s1,s2 or s3 is no more than 8.

When you put a ‘=’ between s2 and s3, and put a operator( ‘+’,’-‘, ‘×’ or ‘÷’.) between s1 and s2, and replace every capital letter with a digit, you get a equation.

You should figure out the number of solutions making the equation right.

Please note that same letters must be replaced by same digits, and different letters must be replaced by different digits. If a number in the equation is more than one digit, it must not have leading zero.

 

Output

For each test case, print an integer in a line. It represents the number of solutions.

 

Sample Input

2A A ABCD BCD B

 

Sample Output

5
72
#include<stdio.h>#include<string.h>#include<map>#include<algorithm>using namespace std;char s1[12], s2[12], s3[12];int a, b, c;int len1, len2, len3;int ans, num;bool vis[12];map<char, int> mm; // 表示映射，int表示 字母char是第几个加入map里面的，从1开始（去除重复）int d[13]; //记录搜索后各字母所对应的值。void dfs(int dep){    int i, j;    if(dep == num)    {        a = b = c = 0;        for(i = 0; i < len1; i++)            a = a * 10 + d[mm[s1[i]]];        for(i = 0; i < len2; i++)            b = b * 10 + d[mm[s2[i]]];        for(i = 0; i < len3; i++)            c = c * 10 + d[mm[s3[i]]];    //printf("%d %d %d\n", a, b, c);        if(a + b == c) ans++;        if(a == b * c && b != 0) ans++;        if(a - b == c) ans++;        if(a * b == c) ans++;                return ;    }    for(i = 0; i <= 9; i++)    {        if(vis[i]) continue;        if(dep == mm[s1[0]] && i == 0 && len1 >= 2) continue; // 搜到最高位数字 每个数前面不能是0，去掉这些情况        if(dep == mm[s2[0]] && i == 0 && len2 >= 2) continue;        if(dep == mm[s3[0]] && i == 0 && len3 >= 2) continue;        vis[i] = 1;        d[dep] = i;        dfs(dep+1);        vis[i] = 0;    }}int main(){    int i, j, cas;    scanf("%d", &cas);    while(cas--)    {        num = 1; ans = 0;        scanf("%s%s%s", s1, s2, s3);        len1 = strlen(s1);        len2 = strlen(s2);        len3 = strlen(s3);        mm.clear();        for(i = 0; i < len1; i++)            if(mm[s1[i]] == 0)     mm[s1[i]] = num++;        for(i = 0; i < len2; i++)            if(mm[s2[i]] == 0) mm[s2[i]] = num++;        for(i = 0; i < len3; i++)            if(mm[s3[i]] == 0) mm[s3[i]] = num++;        memset(vis, 0, sizeof(vis));        dfs(1);        printf("%d\n", ans);    }    return 0;}