#295(div.2) C.DNA Alignment

标签：数学

1.题目描述：点击打开链接

2.解题思路：比赛时没有想到好的思路，后来才发现，只需要t串中的字符是s串中出现次数最多的字符即可，根据乘法原理可知：最终结果是pow(num,n)，其中num是s串中次数最多的字符的个数。

3.代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

int n;
const int MOD = 1000000000 + 7;
string s;
int vis[4];
char*cg = "ACGT";
int id(char c){ return strchr(cg, c) - cg; }

int P(int e, int n)
{
	if (n == 0)return 1;
	int x = P(e, n / 2);
	long long ans = (long long)x*x%MOD;
	if (n & 1)ans = ans*e%MOD;
	return (int)ans;
}
int main()
{
	//freopen("test.txt", "r", stdin);
	while (cin >> n)
	{
		cin >> s;
		int cnt = 0;
		int num = 0;
		int len = s.length();
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < len; i++)
		{
			vis[id(s[i])]++;
			cnt = max(cnt, vis[id(s[i])]);
		}
		for (int i = 0; i < 4;i++)
		if (vis[i] == cnt)
			num++;
		cout << P(num, n) << endl;
	}
	return 0;
}

#295(div.2) C.DNA Alignment

标签：数学

原文地址：http://blog.csdn.net/u014800748/article/details/44022795