#C++初学记录（ACM8-6-cf-f题）
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F. Vanya and Label
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 1e9?+?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 1e9?+?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
正确代码
#include<bits/stdc++.h>
using namespace std;
#define modd 1000000007
int main()
char a[100006];
cin>>a;
long long lena=strlen(a);
long long ans=1;
long long n;
for(int i=0;i<lena;i++)
n=0;
if(a[i]>='0'&&a[i]<='9')
n=a[i]-'0';
if(a[i]>='A'&&a[i]<='Z')
n=a[i]-'A'+10;
if(a[i]>='a'&&a[i]<='z')
n=a[i]-'a'+36;
if(a[i]=='-')
n=62;
if(a[i]=='_')
n=63;
for(int j=0;j<6;j++)
if(((n>>j)&1)==0)
ans=(ans*3)%modd;
cout<<ans<<endl;
return 0;
题目理解
给出一个字符串，字符串中可能含有1-9的数字，a-z的字母，A-Z的字母，然后将他们全部转化成题目给出的编码，得到编码后，通过程序进行判断有多少个字符组合通过AND（&）符运算后不会改变字符串中字符的编码。
相关知识点
本题完成需要读懂题目，判断字符串中编码不仅仅是将其化为ascll码，而是要根据题意进行初始编码后再进行化为二进制，其次二进制的判断是位移运算符与AND运算符结合完成的，具体代码为
for(int j=0;j<6;j++)
if(((n>>j)&1)==0)
ans=(ans*3)%modd;
由题意得出最大代码为63的“_”，即不超过64位，因此二进制可以化为六位数正好小于2^6，到此代码编写与理解完成。
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