D. Yet Another Subarray Problem（枚举）

最新推荐文章于 2021-09-01 09:36:25 发布

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给定数组a和两个整数m、k，计算数组中子序列的最大成本，成本定义为子序列元素之和减去子序列长度除以m的上取整倍数的k。寻找并返回最大成本。

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D. Yet Another Subarray Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array a1,a2,…,ana1,a2,…,an and two integers mm and kk.

You can choose some subarray al,al+1,…,ar−1,aral,al+1,…,ar−1,ar.

The cost of subarray al,al+1,…,ar−1,aral,al+1,…,ar−1,ar is equal to ∑i=lrai−k⌈r−l+1m⌉∑i=lrai−k⌈r−l+1m⌉, where ⌈x⌉⌈x⌉ is the least integer greater than or equal to xx.

The cost of empty subarray is equal to zero.

For example, if m=3m=3, k=10k=10 and a=[2,−4,15,−3,4,8,3]a=[2,−4,15,−3,4,8,3], then the cost of some subarrays are:

a3…a3:15−k⌈13⌉=15−10=5a3…a3:15−k⌈13⌉=15−10=5;
a3…a4:(15−3)−k⌈23⌉=12−10=2a3…a4:(15−3)−k⌈23⌉=12−10=2;
a3…a5:(15−3+4)−k⌈33⌉=16−10=6a3…a5:(15−3+4)−k⌈33⌉=16−10=6;
a3…a6:(15−3+4+8)−k⌈43⌉=24−20=4a3…a6:(15−3+4+8)−k⌈43⌉=24−20=4;
a3…a7:(15−3+4+8+3)−k⌈53⌉=27−20=7a3…a7:(15−3+4+8+3)−k⌈53⌉=27−20=7.

Your task is to find the maximum cost of some subarray (possibly empty) of array aa.

Input

The first line contains three integers nn, mm, and kk (1≤n≤3⋅105,1≤m≤10,1≤k≤1091≤n≤3⋅105,1≤m≤10,1≤k≤109).

The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109).

Output

Print the maximum cost of some subarray of array aa.

Examples

input

Copy

7 3 10
2 -4 15 -3 4 8 3

output

Copy

input

Copy

5 2 1000
-13 -4 -9 -20 -11

output

Copy

由表达式可知最后权值中每隔m个就会减k

那么我们对原数组进行处理每隔m个减去k

枚举又端点modm的数即可

#include<bits/stdc++.h>
using namespace std;
int a[300005];
int b[300005];
int main()
{
    int n,m,k;
    scanf("%d%d%d",&n,&m,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",a+i);
    }
    long long MAX=0;
    for(int j=0;j<m;j++)
    {
        for(int i=1;i<=n;i++)
        {
            b[i]=a[i];
            if(i%m==j) b[i]-=k;
        }
        long long sum=0;
        for(int i=1;i<=n;i++)
        {
            sum+=b[i];
            if(i%m==j) MAX=max(sum,MAX);
            sum=max(sum,0LL);
        }
    }
    printf("%lld\n",MAX);
}