题目链接：

PKU:

HDU:

Description

A wild number is a string containing digits and question marks (like 36?1?8). A number X matches a wild number W if they have the same length, and every non-question mark character in X is equal to the character in the same position in W (it means that you can replace a question mark with any digit). For example, 365198 matches the wild number 36?1?

8, but 360199, 361028, or 36128 does not. Write a program that reads a wild number W and a number X from input, both of length n, and determines the number of n-digit numbers that match W and are greater than X.

Input

There are multiple test cases in the input. Each test case consists of two lines of the same length. The first line contains a wild number W, and the second line contains an integer number X. The length of input lines is between 1 and 10 characters. The last line of input contains a single character #.

Output

For each test case, write a single line containing the number of n-digit numbers matching W and greater than X (n is the length of W and X).

Sample Input

36?

1?

8 236428 8?3 910 ? 5 #

Sample Output

10004

Source

题意：

给出两个数字字符串，串一中有'?',问在‘？’位置填数字共同拥有多少中填法，保证串一大于串二！

代码例如以下：

#include 
     
      #include 
      
       #include 
       
        const int maxn = 17;typedef __int64 LL;int cont;LL ans;char s1[maxn];char s2[maxn];LL POW(LL n, LL k){    LL s = 1;    for(LL i = 0; i < k; i++)    {        s*=n;    }    return s;}void solve(int len, int k){    for(int i = 0; i < len; i++)    {        if(s1[i] == '?
       
      
     

') { ans+=(9-(s2[i]-'0'))*POW(10,--cont); } else if(s1[i] < s2[i]) return ; else if(s1[i] > s2[i]) { ans+=POW(10,cont); return ; } } } int main() { while(~scanf("%s",s1)) { ans = 0; cont = 0; if(s1[0] == '#') break; scanf("%s",s2); int len = strlen(s1); for(int i = 0; i < len; i++) { if(s1[i] == '?') cont++; } solve(len,0); printf("%I64d\n",ans); } return 0; }