C. DNA Alignment
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya became interested in bioinformatics. He’s going to write an article about similar cyclic DNA sequences, so he invented a new method for determining the similarity of cyclic sequences.
Let’s assume that strings s and t have the same length n, then the functionh(s,t) is defined as the number of positions in which the respective symbols ofs and t arethe same. Function h(s,t) can be used to define the function of Vasya distanceρ(s,t):
where is obtained from strings, by applying left circular shift i times. For example,ρ("AGC","CGT")=h("AGC","CGT")+h("AGC","GTC")+h("AGC","TCG")+h("GCA","CGT")+h("GCA","GTC")+h("GCA","TCG")+h("CAG","CGT")+h("CAG","GTC")+h("CAG","TCG")=1+1+0+0+1+1+1+0+1=6
Vasya found a string s of length n on the Internet. Now he wants to count how many strings t there are such that the Vasya distance from the string s attains maximum possible value. Formally speaking, t must satisfy the equation: .
Vasya could not try all possible strings to find an answer, so he needs your help. As the answer may be very large, count the number of such strings modulo109+7.
Input
The first line of the input contains a single integer n (1≤n≤105).
The second line of the input contains a single string of length n, consisting of characters "ACGT".
Output
Print a single number— the answer modulo 109+7.
Sample test(s)
Input
1C
Output
1
Input
2AG
Output
4
Input
3TTT
Output
1
Note
Please note that if for two distinct strings t1 andt2 values ρ(s,t1) и ρ(s,t2) are maximum among all possiblet, then both strings must be taken into account in the answer even if one of them can be obtained by a circular shift of another one.
In the first sample, there is ρ("C","C")=1, for the remaining stringst of length 1 the value of ρ(s,t) is 0.
In the second sample, ρ("AG","AG")=ρ("AG","GA")=ρ("AG","AA")=ρ("AG","GG")=4.
In the third sample, ρ("TTT","TTT")=27
题意：
两个字符串，，p值为将将两个字符串最右面的字符移到最左边的所有可能的h值的和，当两个字符串有相同的，取1，否则取0，现给出一个字符串，求所有p值取最大值的另一个字符串。
题解：
首先要明白，由于是可以移动位置的，所以要想使p值最大，则另一个字符串只可能出现出现次数最多的字符。
代码：
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn=100000+1000;const int mod=1000000000+7;char s[maxn];int n;int a[5];long long ans;long long pow_q(long long x,int n){ans=1;while(n){if(n&1)ans=(ans*x)%mod;x=(x*x)%mod;n=n>>1;}return ans;}int main(){while(~scanf("%d",&n)){scanf("%s",s);memset(a,0,sizeof(a));for(int i=0;i<n;i++){if(s[i]=='A')a[0]++;if(s[i]=='C')a[1]++;if(s[i]=='G')a[2]++;if(s[i]=='T')a[3]++;}sort(a,a+4);int maxc=a[3];int cou=0;for(int i=0;i<4;i++)if(a[i]==maxc)cou++;long long ans;ans=pow_q(cou,n);cout<<ans<<endl;}return 0;}
游手好闲会使人心智生锈
