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  • D. Yet Another Subarray Problem(枚举)

    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

    7
    

    input

    Copy

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

    output

    Copy

    0
    

    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
    ",MAX);
    }
    
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  • 原文地址:https://www.cnblogs.com/caowenbo/p/11852242.html
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