CodeForces 1197D Yet Another Subarray Problem

Time limit 2000 ms

Memory limit 262144 kB

Source Educational Codeforces Round 69 (Rated for Div. 2)

Tags dp greedy math *1900

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官方题解

At first let's solve this problem when )=0.

We will calculate it the following way. maxli will be the maximum of two values:

• 0);
• i−1.

The maximum sum of some subarray is equal to 1≤i≤nmaxli.

So, now we can calculate the values of ) the same way.

besti is the maximum of two values:

• 0;
• i−m.

After calculating all values )−k.

源代码

#include<stdio.h>
#include<algorithm>

int n,m,k;
long long a[300010];
long long dp[300010],ans;
int main()
{
	//freopen("test.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for(int i=1;i<=n;i++) scanf("%lld",a+i),a[i]+=a[i-1];
	for(int i=1;i<=n;i++)
	{
		for(int j=i;j+m>=i;j--)
			dp[i]=std::max(dp[i],a[i]-a[j]);
		dp[i]-=k;
		dp[i]=std::max(0LL,dp[i]);
		if(i>m) dp[i]=std::max(dp[i],dp[i-m]+a[i]-a[i-m]-k);
		ans=std::max(dp[i],ans);
	}
	printf("%lld\n",ans);
	return 0;
}