Codeforces Round #355 (Div. 2) C. Vanya and Label

C. Vanya and Label

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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 

题意：给你一个字符串 s ，让你找另两个字符串，使得两个字符串对应位置的字符进行&运算得出的字符为 s 当前位置的字符，求有多少种方案

思路：单个字符的&运算的方案可以求出来，因为字符串中字符间没有联系，所以说将每个字符的方案数相乘取模即可。

ac代码：

/* ***********************************************
Author       : AnICoo1
Created Time : 2016-08-22-10.33 Monday
File Name    : D:\MyCode\2016-8月\2016-8-22.cpp
LANGUAGE     : C++
Copyright  2016 clh All Rights Reserved
************************************************ */
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define ll __int64
#define mem(x,y) memset(x,(y),sizeof(x))
#define PI acos(-1)
#define gn (sqrt(5.0)+1)/2
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
const ll INF=1e9+10;
const int MAXN=1e6+10;
const int MOD=1e9+7;
//head
char str[MAXN];

int get(char ch)
{
    if(ch>='0'&&ch<='9')
        return ch-'0';
    else if(ch>='a'&&ch<='z')
        return ch-'a'+36;
    else if(ch>='A'&&ch<='Z')
        return ch-'A'+10;
    else if(ch=='-')
        return 62;
    else if(ch=='_')
        return 63;
}

int main()
{
    scanf("%s",str);
    int len=strlen(str);
    ll ans=1;
    for(int i=0;iint num=get(str[i]);ll cnt=0;
        for(int j=0;j<=63;j++)
        {
            for(int k=0;k<=63;k++)
            {
                int q=k&j;
                if(q==num)
                    cnt++;
            }
        }
        ans=(ans*cnt)%MOD;
    }
    printf("%I64d\n",ans);
    return 0;
}