Codeforces 521A DNA Alignment Law

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Test instructions

A string s that is given a length of N.

Constructs a string t with a length of N. Makes the P (s,t) value the largest, asking how many different t

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

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 Ideas:

First we construct a letter of T, then the letter and the letter s of any one of the letters in any two positions will be matched once, and the corresponding number of times have n times. So the construction of this letter to the answer contribution is num[this_letter] * n

If a is filled in T, then the value of P (s,t) increases (the number of a letters in s) *n

So in order for p to be the largest, the letters that can be filled in T must be the largest number of letters in S.

The number of letters in S is as many as possible, and how many letters can be filled in each position in T.

#include <cstdio> #include <cstring> #include <algorithm>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\n", ans);    return 0;}

Codeforces 521A DNA Alignment Law