Codeforces Round #295 (Div. 2) C. DNA Alignment(数学)

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题目:

http://codeforces.com/problemset/problem/520/C

题意:

长度为n 的字符串, 与只含有"ACGT"的n长度的字符串作比较,求出得到得数最大的字符串有几条.

思路:

求出字符串中重复次数最多的字母的个数x , 则答案为 ans = x ^ n.

AC.

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int mod = 1e9+7;char s[100005];int a[5];long long cal(long long x, int n){    long long ans = 1;    while(n > 0) {        if(n & 1) ans = (ans * x) % mod;        x = (x * x) % mod;        n >>= 1;    }    return ans;}int main(){//freopen("in", "r", stdin);    int n;    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 maxn = a[3];        long long cnt = 0;        for(int i = 0; i < 4; ++i) {            if(a[i] == maxn) cnt++;        }        long long ans = cal(cnt, n);        printf("%I64d\n", ans);    }    return 0;}

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