# coreBugZJ

## A hard Aoshu Problem， ACM/ICPC 2010/2011 亚洲，福州区域赛 J， UVA 5107

 5107 - A hard Aoshu ProblemAsia - Fuzhou - 2010/2011

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   = 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 `+', `-`, ` x' 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 (``+`','`-``, ` x' 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

```572
Fuzhou 2010-2011
```

1 #include <stdio.h>
2 #include <string.h>
3
4 typedef long long Lint;
5
6 #define  L  10
7
8 char s1[ L ], s2[ L ], s3[ L ];
9
10 #define  DIGIT  6
11
12 int digit[ DIGIT ], used[ DIGIT ];
13 int a[ L ], b[ L ], c[ L + L ];
14
15 void parse( char s[], int a[], int size ) {
16         int i, len = strlen( s );
17         memset( a, 0, size );
18         for ( i = 0; i < len; ++i ) {
19                 used[ a[ len - i - 1 ] = s[ i ] - 'A' + 1 ] = 1;
20         }
21 }
22
23 void init() {
24         memset( used,  0sizeof(used)  );
25         parse( s1, a, sizeof(a) );
26         parse( s2, b, sizeof(b) );
27         parse( s3, c, sizeof(c) );
28 }
29
30 int parseInt( int a[], int n, Lint *num ) {
31         int i = n-1;
32         while ( (i>=0&& (a[i]==0) ) --i;
33         if ( (i<0|| ((i>0)&&(digit[a[i]]==0)) ) return 0;
34         *num = 0;
35         for ( ; i >= 0--i ) {
36                 if ( digit[ a[ i ] ] > 9 ) return 0;
37                 *num = (*num) * 10 + digit[ a[ i ] ];
38         }
39         return 1;
40 }
41
42 int equal() {
43         Lint x, y, z;
44         int i, ans = 0;
45         for ( i = 1; i < DIGIT; ++i ) {
46                 if ( (digit[i]<10&& (used[i]==0) ) {
47                         return 0;
48                 }
49         }
50         if ( !parseInt( a, sizeof(a)/sizeof(a), &x ) ) return 0;
51         if ( !parseInt( b, sizeof(b)/sizeof(b), &y ) ) return 0;
52         if ( !parseInt( c, sizeof(c)/sizeof(c), &z ) ) return 0;
53         if ( x + y == z ) ++ans;
54         if ( x - y == z ) ++ans;
55         if ( x * y == z ) ++ans;
56         if ( (y!=0&& (x%y==0&& (x/y==z) ) ++ans;
57         return ans;
58 }
59
60 int solve() {
61         int ans = 0, have[ 22 ] = { 0 };
62 #define  F(i) for ( digit[ i ] = 0; digit[ i ] < 11; ++digit[ i ] ) { \
63                 if ( have[ digit[ i ] ] ) continue; \
64                 if ( digit[ i ] < 10 ) have[ digit[ i ] ] = 1;
65 #define  E(i)  have[ digit[ i ] ] = 0; \
66         }
67         F(1) F(2) F(3) F(4) F(5) {
68                 ans += equal();
69         } E(5) E(4) E(3) E(2) E(1)
70         return ans;
71 }
72
73 int main() {
74         int td;
75         scanf( "%d"&td );
76         while ( td-- > 0 ) {
77                 scanf( "%s%s%s", s1, s2, s3 );
78                 init();
79                 printf( "%d\n", solve() );
80         }
81         return 0;
82 }
83

posted on 2011-03-24 22:14 coreBugZJ 阅读(1974) 评论(3)  编辑 收藏 引用 所属分类: ACM @Bill