Codeforces Round #642 (Div. 3) E—K-periodic Garland dp

#include <bits/stdc++.h>
using namespace std;
const int N= 1e6 + 5;
char a[N];
int dp[N],s[N];
int main()
{
	int t;
	scanf("%d", &t);
	while(t--)
	{
		int n, k;
		scanf("%d%d", &n, &k);
		scanf("%s", a + 1);
		dp[0]=dp[1]=0;
		s[0]=0;
		for(int i = 1; i <= n; i++)
		{
			//1的个数
			s[i] = s[i - 1] + (a[i] == '1');
			//把前面的1都抹掉
			dp[i] = s[i - 1];
			//如果大于等于k，默认当前位置是1
			if(i >= k)
				//				上一个1的位置
				dp[i] = min(dp[i], dp[i - k] + s[i - 1] - s[i - k]);
			//如果是1，再加上代价
			if(a[i] == '0')
				dp[i]++;
		}
		int ans = 1e9;
		//只是到i，把i后面的1都抹掉
		for(int i = 0; i <= n; i++)
			ans = min(ans, dp[i] + s[n] - s[i]);
		printf("%d\n", ans);
	}

	return 0;
}

标签：Garland,int,scanf,d%,Codeforces,char,642,dp
来源： https://www.cnblogs.com/QingyuYYYYY/p/12902324.html