[暴力枚举]Codeforces 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 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.
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 '_'.
Print a single integer — the number of possible pairs of words, such that their bitwise AND is equal to string s modulo 109 + 7.
z
3
V_V
9
Codeforces
130653412
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，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s
思路:看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int amn=1e5+5,mod=1e9+7; 4 char a[amn]; 5 int main(){ 6 ios::sync_with_stdio(0); 7 cin>>a; 8 long long ans=1,sum; 9 int len=strlen(a),in,jg; 10 for(int i=0;i<len;i++){ 11 if(a[i]>='0'&&a[i]<='9')in=a[i]-'0'; 12 else if(a[i]>='A'&&a[i]<='Z')in=a[i]-'A'+10; 13 else if(a[i]>='a'&&a[i]<='z')in=a[i]-'a'+36; 14 else if(a[i]=='-')in=62; 15 else in=63; 16 sum=0; 17 for(int j=0;j<=63;j++){ 18 for(int k=0;k<=63;k++){ 19 jg=j&k; 20 if(jg==in)sum++; 21 } 22 } 23 ans=((ans%mod)*(sum%mod))%mod; 24 } 25 printf("%lld\n",ans); 26 } 27 /*** 28 给你一个字符串s，把字符映射为一些数字，问有多少对和这个字符串长度相同的字符串逐位AND得到仍能得到字符串s 29 看每位有多少种可能，再把他们相乘并取模，就是答案，注意到最多只有64个字符，所以O(n*n)暴力，4096*1e5大概在4e8可以在一秒内跑完 30 ***/
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