Vanya and Label - codeforces - problem/677/C

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.

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.

In the first sample, there are 3 possible solutions:

z&_ = 61&63 = 61 = z
_&z = 63&61 = 61 = z
z&z = 61&61 = 61 = z

题目链接

http://codeforces.com/problemset/problem/677/C

题目大意

每个字符对应一个数字，已知一个字符串，问存在多少对字符串进行 ‘&’ 计算后能得到已知字符串，答案mod1e9 + 7

数据范围

s (1 ≤ |s| ≤ 100 000)
modulo 1e9 + 7

解题思路

按位与 ： 0 & 1 = 0；0 & 0 = 0； 1 & 1 = 1；1 & 0 = 0；
由此可知，已知的数如果二进制为1，就只有一种情况 1 & 1；
已知的数如果二进制为0，则有3种情况
综上所述，答案就是判断已知数的二进制有多少个0，将0的个数 * 3 mod 1e9 + 7
注意
因为字符对应范围是（0 ~ 63）
64是2的6次幂，而63转换成二进制就是5位，有效数是5位，位数不足要用0补

解决代码

#include
#include
#include
#include
using namespace std;
typedef long long ll;
int mod = 1e9 + 7;
int getnum(char a)
{
if(a >= '0'&&a <= '9') return a - '0';
if(a >= 'A'&& a <= 'Z') return a - 'A' + 10;
if(a >= 'a' && a <= 'z')return a - 'a' + 36;
if(a == '-') return 62;
if(a == '_') return 63;
}
int main()
{
string a;
ll ans = 1;
cin >> a;
for(int i = 0;i < a.size();i++){
	int aa = getnum(a[i]);
	for(int j = 0;j < 6;j++){
		if(!((aa >> j)& 1)) ans = ans * 3 % mod;
		}
	}
	printf("%lld\n",ans);
return 0;
}