Codefoces 677C Vanya and Label (预处理/翻译)

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.

题目大意：

找一个与输入串字符相同的串。。。使每一个能对应的字符与 输入的 对应的字符 取和（&） 且必须还点与原串的对应的值相等。

V_V            V代表 31 二进制 011111          _代表63 二进制1111111

011111&011111 111111&011111 011111&111111 

111111&111111

一共是3*3

V    _     V

V    V     _

1.V_V   V_V            2.V__  V_V              3. V_V  V__

4.__V    V_V             5.___ V_V              6.__V   V__

7.V_V    __V           8.V__ __V                9.V_V ___

#include<stdio.h>#include<string.h>using namespace std;const int MOD=1e9+7;char s[100005];int sum[65];int change(char c){    if(c>='0'&&c<='9')return c-'0';    else if(c>='A'&&c<='Z')return c-'A'+10;    else if(c>='a'&&c<='z')return c-'a'+36;    else if(c=='-')return 62;    else if(c=='_')return 63;}int main(){    for(int i=0;i<64;i++)    {        for(int j=0;j<64;j++)        {            sum[i&j]++;        }    }    long long ans=1;    scanf("%s",s);    for(int i=0;i<strlen(s);i++)    {        ans=((ans*sum[change(s[i])]))%MOD;    }    printf("%d\n",ans);}

预处理所以   i&j    再计算。