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.
with
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
Meaning of the questions: to give you a string s, the character is mapped to some figures, and asked how many of the same string length of the string-by-bit AND get still get the string s
Thinking: Look how many of each species could, then they multiplied and modulo, is the answer, noting that only a maximum of 64 characters, so O (n * n) Violence, 4096 * 1e5 probably within one second at 4e8 finish
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 is ANS = ((% MOD ANS) * (MOD SUM%))% MOD; 24 } 25 the printf ( " % LLD \ n- " , ANS); 26 is } 27 / * ** 28 give you a string s, the character mapping for a number of figures, and asked how many of the same string length of the string-by-bit aND get still get the string S 29 to see how many of each species could, then they multiplied and modulo, is the answer , noting that only a maximum of 64 characters, so O (n * n) violence, 4096 * 1e5 probably can finish within one second in 4E8 30 ** * /