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:
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
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