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

Editorial Announcement (en) Tutorial #1 (en) Tutorial #2 (en) Tutorial #3 (ru)

官方题解

At first let's solve this problem when m=1" role="presentation">m=1m=1 and k=0" role="presentation">k=0k=0 (it is the problem of finding subarray with maximum sum). For each position from 1" role="presentation">11 to n" role="presentation">nn we want to know the value of maxli=max1≤j≤i+1sum(j,i)" role="presentation">maxli=max1≤j≤i+1sum(j,i)maxli=max1≤j≤i+1sum(j,i), where sum(l,r)=∑k=lk≤rak" role="presentation">sum(l,r)=∑k=lk≤raksum(l,r)=∑k=lk≤rak, and sum(x+1,x)=0" role="presentation">sum(x+1,x)=0sum(x+1,x)=0.

We will calculate it the following way. maxli" role="presentation">maxlimaxli will be the maximum of two values:

• 0" role="presentation">00 (because we can take segments of length 0" role="presentation">00);
• ai+maxli−1" role="presentation">ai+maxli−1ai+maxli−1.

The maximum sum of some subarray is equal to max1≤i≤nmaxli" role="presentation">max1≤i≤nmaxlimax1≤i≤nmaxli.

So, now we can calculate the values of besti=max0≤len,i−len⋅m≥0(sum(i−len⋅m+1,i)−len∗k)" role="presentation">besti=max0≤len,i−len⋅m≥0(sum(i−len⋅m+1,i)−len∗k)besti=max0≤len,i−len⋅m≥0(sum(i−len⋅m+1,i)−len∗k) the same way.

besti" role="presentation">bestibesti is the maximum of two values:

• 0;
• sum(i−m+1,i)−k+besti−m" role="presentation">sum(i−m+1,i)−k+besti−msum(i−m+1,i)−k+besti−m.

After calculating all values besti" role="presentation">bestibesti we can easily solve this problem. At first, let's iterate over the elements besti" role="presentation">bestibesti. When we fix some element besti" role="presentation">bestibesti, lets iterate over the value len=1,2,…,m" role="presentation">len=1,2,…,mlen=1,2,…,m and update the answer with value besti+sum(i−len,i−1)−k" role="presentation">besti+sum(i−len,i−1)−kbesti+sum(i−len,i−1)−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;
}