华为oj DNA序列

很简单直接找哪5个连续的字串中GC出现的次数高就行

#include<iostream>
using namespace std;
int main()
{
	char in[100] ;
	int start = 0, end = 0, num, len, finalstart, finalend, tempsum = 0, sum = 0;
	gets_s(in);
	len = strlen(in);
	cin >> num;
	end = num - 1;
	for (; end<len; start++, end++)
	{
		for (int i = start; i <= end; i++)
		{
			if (in[i] == 'G' || in[i] == 'C')
			{
				tempsum++;
			}
		}
		if (tempsum>sum)
		{
			sum = tempsum;
			finalstart = start;
			finalend = end;
		}
		tempsum = 0;
	}
	for (int i = finalstart; i <= finalend; i++)
	{
		cout << in[i];
	}
	cout << endl; 
	return 0;
}

结论：start和end的距离始终是5，所以这种解法的时间复杂度是O（n）