Codeforces 521A DNA Alignment Law

Codeforces 521A DNA Alignment Law

 

Question:

Given a string of n s.

Construct a string t with a length of n. So that the p (s, t) value is the largest, ask how many different t

H (s, t) = number of identical letters in the corresponding position

 

P ( AGC, Bytes, CGT) Bytes = bytes H( AGC, Bytes, CGT) Accept + accept H( AGC, Bytes, GTC) Accept + accept H( AGC, Bytes, TCG) Accept + accept H( GCA, Bytes, CGT) Accept + accept H( GCA, Bytes, GTC) Accept + accept H( GCA, Bytes, TCG) Accept + accept H( CAG, Bytes, CGT) Accept + accept H( CAG, Bytes, GTC) Accept + accept H( CAG, Bytes, TCG) Bytes = bytes 1 worker + WORKER 1 worker + worker 0 worker + worker 0 worker + WORKER 1 worker + WORKER 1 worker + WORKER 1 worker + worker 0 worker + WORKER 1 worker = worker 6 Ideas:

 

First, we construct a Letter of t, so that the letter and any letter of s will match at any two locations, and the corresponding times are n times. Therefore, the contribution of the constructed letter to the answer is Num [this_Letter] * n.

If A is entered in t, the value of p (s, t) is increased (the number of letters A in s) * n

To maximize p, the number of letters that can be entered in t must be the one with the largest number of letters in s.

The maximum number of letters in s and the number of letters that can be entered at each position in t.

 

#include 
 
  #include 
  
   #include using namespace std;const int MAX_N = 100007;const long long mod = 1000000007;long long Pow(long long x, long long n) {    long long res = 1;    while (n > 0) {        if (n & 1) res = res * x % mod;        x = x * x % mod;        n >>= 1;    }    return res;}char str[MAX_N];int a[5], n;int main() {    scanf(%d%s, &n, str);    for (int i = 0; str[i]; ++i) {        if (str[i] == 'A') ++a[0];        else if (str[i] == 'C') ++a[1];        else if (str[i] == 'G') ++a[2];        else ++a[3];    }    int up = 0;    for (int i = 0; i < 4; ++i) up = max(up, a[i]);    int cnt = 0;    for (int i = 0; i < 4; ++i) if (a[i] == up) ++cnt;    long long ans = Pow(cnt, n);    printf(%I64d, ans);    return 0;}