C. Vanya and Label

题目连接：

http://www.codeforces.com/contest/677/problem/C

Description

While walking down the street Vanya saw a label “Hide&Seek”. Because he is a programmer, he used & as a bitwise AND for these two words represented as a integers in base 64 and got new word. Now Vanya thinks of some string s and wants to know the number of pairs of words of length |s| (length of s), such that their bitwise AND is equal to s. As this number can be large, output it modulo 109 + 7.

To represent the string as a number in numeral system with base 64 Vanya uses the following rules:

digits from ‘0’ to ‘9’ correspond to integers from 0 to 9;
letters from ‘A’ to ‘Z’ correspond to integers from 10 to 35;
letters from ‘a’ to ‘z’ correspond to integers from 36 to 61;
letter ‘-‘ correspond to integer 62;
letter ‘_’ correspond to integer 63.

Input

The only line of the input contains a single word s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters ‘-‘ and ‘_’.

Output

Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 109 + 7.

Sample Input

Codeforces

Sample Output

130653412

Hint

题意

给你一个字符串，问你有多少对相同长度的字符串 & 起来之后，恰好等于这个字符串

这个字符串的每个字符都是代表着0-63之间的数字

题解：

首先每个字符是独立的，我们把每个字符的方案数知道，然后再全部乘起来就好了

0-63是 2^6，那么我们就按位去考虑就好了

如果对于这一位是0的话，那么就有3种方案0&1,1&0,0&0，如果这一位为1的话，只有一种方案

所以看一共有多少个位是0就好了，然后3的那么多次幂就行了

代码

#include<bits/stdc++.h>
using namespace std;
const int mod = 1e9+7;
string s;
int getidx(char c)
{
    if(c>='0'&&c<='9')return c-'0';
    if(c>='A'&&c<='Z')return c-'A'+10;
    if(c>='a'&&c<='z')return c-'a'+36;
    if(c=='-')return 62;
    if(c=='_')return 63;
}
int main()
{
    cin>>s;
    long long ans = 1;
    for(int i=0;i<s.size();i++)
    {
        int p = getidx(s[i]);
        for(int j=0;j<6;j++)
            if(!((p>>j)&1))
                ans=ans*3%mod;
    }
    cout<<ans<<endl;
}