| 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
/*
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=;
for(int i=;i<=b;i++)
ans*=;
return ans;
}
int main()
{
//freopen("in.txt","r",stdin);
char a[N],b[N];
while(scanf("%s",&a)!=EOF&&strcmp(a,"#")!=)
{
long long ans=;//计时器
int sum=,cur=;//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=;a[i];i++)
if(a[i]=='?')
sum++;
for(int i=;a[i];i++)
{
if(a[i]!='?')
{
if(a[i]>b[i])
{
ans+=power(,sum-cur);
break;
}
else if(a[i]<b[i])
break;
}
else if(a[i]=='?')
{
cur++;
ans+=power(,sum-cur)*(-b[i]+'');
}
}
printf("%lld\n",ans);
}
return ;
}