poj 3340 Barbara Bennett's Wild Numbers(数位DP)

Barbara Bennett's Wild Numbers

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3153		Accepted: 1143

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

100
0
4

Source

Tehran 2006

/*
回来的第一道题，几天不用脑子生锈了
*/
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define N 20
using namespace std;
long long power(long long a,long long b)//类似于pow函数的一个功能，因为颇为太小了
{
    long long ans=1;
    for(int i=1;i<=b;i++)
        ans*=10;
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    char a[N],b[N];
    while(scanf("%s",&a)!=EOF&&strcmp(a,"#")!=0)
    {
        long long ans=0;//计时器
        int sum=0,cur=0;//计算问号的次数和枚举目前为止出现问号的次数
        scanf("%s",&b);
        for(int i=0;a[i];i++)
            if(a[i]=='?')
                sum++;
        for(int i=0;a[i];i++)
        {
            if(a[i]!='?')
            {
                if(a[i]>b[i])
                {
                    ans+=power(10,sum-cur);
                    break;
                }
                else if(a[i]<b[i])
                    break;
            }
            else if(a[i]=='?')
            {
                cur++;
                ans+=power(10,sum-cur)*(9-b[i]+'0');
            }
        }
        printf("%lld
",ans);
    }
    return 0;
}