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
思路：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
#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;
}
C. Vanya and Label
time limit per test
memory limit per test
input
output
& 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.
64
- 0' to '9' correspond to integers from 0 to 9;
- A' to 'Z' correspond to integers from 10 to 35;
- a' to 'z' correspond to integers from 36 to 61;
- -' correspond to integer 62;
- _' correspond to integer 63.
Input
s (1 ≤ |s| ≤ 100 000), consisting of digits, lowercase and uppercase English letters, characters '-' and '_'.
Output
s modulo 109.
Examples
input
z
output
3
input
V_V
output
9
input
Codeforces
output
130653412
Note
For a detailed definition of bitwise AND we recommend to take a look in the corresponding article in Wikipedia.
3
- z&_ = 61&63 = 61 = z
- _&z = 63&61 = 61 = z
- z&z = 61&61 = 61 = z
