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hdu 3699 10 福州 现场 J - A hard Aoshu 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.
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.
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
572
暴力
#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;}
hdu 3699 10 福州 现场 J - A hard Aoshu Problem
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