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题目链接：
PKU:http://poj.org/problem?id=3340
HDU:http://acm.hdu.edu.cn/showproblem.php?pid=2410
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?
82364288?3910?5#
Sample Output
100 0 4
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 <cstdio>
#include <cstring>
#include <cmath>
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;}
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