codeforces 1197D. Yet Another Subarray Problem 动态规划

题意：给你n个数，还有m和k，让你找一个区间使这个公式的值最大其中⌈x⌉是对x向上取整，l和r是所选的区间的左右下标。

思路：定义一个dp[i][j]，表示第i个数作为右端点，区间长度对m取余为j的最大值。那么就可以得到转移方程：

dp[i][j%m]=max(a[i]-k,dp[i-1][0]+a[i]-k );///当j等于1时
dp[i][j%m]=dp[i-1][(j-1+m)%m]+a[i];///当j不等于1时

余数等于1时就相当于从余数为零再加一个a[i] 那么K也要多减一个,其他时候因为j-1和j 向上取整相同就不需要多减K。

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

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

output

input

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

output

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <vector>
#include <map>
#include <cmath>
using namespace std ;
typedef long long ll;
const ll mod =1e9+7;
const int manx= 3e5+100;
ll a[manx];
ll dp[manx][12];
int main()
{ll n,m,k;scanf("%lld%lld%lld",&n,&m,&k);for(int i=1; i<=n; i++)scanf("%lld",&a[i]);ll ans= 0;for(int i=1; i<=n; i++){for(int j=1; j<=m&&j<=i; j++){if(j==1)dp[i][j%m]=max(a[i]-k,dp[i-1][0]+a[i]-k );elsedp[i][j%m]=dp[i-1][(j-1+m)%m]+a[i];ans=max(dp[i][j%m],ans );}}printf("%lld\n",ans);return 0;
}

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