hdu 3699 10 福州 现场 J - A hard Aoshu Problem 暴力 难度:0
2014-11-10 19:56
411 查看
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
暴力
卡题原因:全排列过后数字是完全打乱的,这时候直接选取新的组合是错误的,必须恢复原来的顺序
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
暴力
卡题原因:全排列过后数字是完全打乱的,这时候直接选取新的组合是错误的,必须恢复原来的顺序
#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);//注意记录原值,否则因为顺序是反的,就不能直接修改取的数字 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; }
相关文章推荐
- HDU 3699 A hard Aoshu Problem(暴力枚举)(2010 Asia Fuzhou Regional Contest)
- HDU 3699 A hard Aoshu Problem (暴力搜索)
- HDU 3699 A hard Aoshu Problem (暴力搜索)
- HDU 3699(J) ——A hard Aoshu Problem(DFS暴力搜索)
- hdu 3695 10 福州 现场 F - Computer Virus on Planet Pandora 暴力 ac自动机 难度:1
- A hard Aoshu Problem HDU - 3699 模拟 map
- A hard Aoshu Problem HDU - 3699
- hdu 3694 10 福州 现场 E - Fermat Point in Quadrangle 费马点 计算几何 难度:1
- HDOJ3699 A hard Aoshu Problem[暴力]
- hdu 3697 10 福州 现场 H - Selecting courses 贪心 难度:0
- HDU 4403 A very hard Aoshu problem (暴力+状态压缩)
- HDU 4403 A very hard Aoshu problem (DFS暴力)
- HDU 4403 A very hard Aoshu problem (DFS暴力)
- HDU 3699 A hard Aoshu Problem
- hdu_4403 A very hard Aoshu problem 暴力搜索
- HDOJ3699 A hard Aoshu Problem[暴力]
- hdu 3696 10 福州 现场 G - Farm Game DP+拓扑排序 or spfa+超级源 难度:0
- HDU 4403 A very hard Aoshu problem(DFS+暴力)
- HDU 4403 A very hard Aoshu problem(DFS)
- hdu 3687 10 杭州 现场 H - National Day Parade 水题 难度:0