POJ 3340 & HDU 2410 Barbara Bennett's Wild Numbers（数学）「建议收藏」

大家好，又见面了，我是全栈君。

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

Tehran 2006

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;}

发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/116429.html原文链接：https://javaforall.cn