A hard Aoshu Problem

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 62768/32768 K (Java/Others)
Total Submission(s): 1620????Accepted Submission(s): 848


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 2 A A A BCD BCD B
Sample Output 5 72

題意：

給三個字符串，只包含A,B,C,D,E, 長度不超過8，對三個字符串中的字符賦值，即相同的字符用0~9之間的字符替換，

s1?( ‘+’,’-‘, ‘×’ or ‘÷’.)? s2=s3，替換后的三個數進行四則運算，要滿足左邊等于右邊，求有幾種解法。

注意：沒有前導 0

#include<stdio.h> #include<string.h> #include<stdlib.h> #include<string> #include<queue> #include<map> using namespace std;int vis[10],flag[10],l,ans; char s1[10],s2[10],s3[10],s[10]; map<char,int>mp,dl; void Cal() {int a=0,b=0,c=0;for(int i=0;i<strlen(s1);i++)a=a*10+mp[s1[i]];for(int i=0;i<strlen(s2);i++)b=b*10+mp[s2[i]];for(int i=0;i<strlen(s3);i++)c=c*10+mp[s3[i]];if(a+b==c) ans++;if(a-b==c) ans++;if(a*b==c) ans++;if(b&&a==b*c) ans++; } void dfs(int num) {if(num>=l) {Cal();return;}for(int i=0;i<10;i++){if(i==0&&dl[s[num]]) continue;if(!flag[i]){flag[i]=1;mp[s[num]]=i;dfs(num+1);flag[i]=0;}} } int main() {int t;scanf("%d",&t);while(t--){memset(vis,0,sizeof(vis));mp.clear();dl.clear();ans=0;scanf("%s%s%s",s1,s2,s3);if(strlen(s1)>1) dl[s1[0]]=1;if(strlen(s2)>1) dl[s2[0]]=1;if(strlen(s3)>1) dl[s3[0]]=1;int i,k,j;l=0;for(i=0;i<strlen(s1);i++){if(vis[s1[i]-'A']) continue;vis[s1[i]-'A']++;s[l++]=s1[i];}for(i=0;i<strlen(s2);i++){if(vis[s2[i]-'A']) continue;vis[s2[i]-'A']++;s[l++]=s2[i];}for(i=0;i<strlen(s3);i++){if(vis[s3[i]-'A']) continue;vis[s3[i]-'A']++;s[l++]=s3[i];}dfs(0);printf("%d\n",ans);}return 0; }

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