Ilya Muromets

?Gym - 100513F

Силачом слыву недаром — семерых одним ударом! From the Russian cartoon on the German fairy tale.

Ilya Muromets is a legendary bogatyr. Right now he is struggling against Zmej Gorynych, a dragon with?n?heads numbered from 1 to?n?from left to right.

Making one sweep of sword Ilya Muromets can cut at most?k?contiguous heads of Zmej Gorynych. Thereafter heads collapse getting rid of empty space between heads. So in a moment before the second sweep all the heads form a contiguous sequence again.

As we all know, dragons can breathe fire. And so does Zmej Gorynych. Each his head has a firepower. The firepower of the?i-th head is?f_i.

Ilya Muromets has time for at most two sword sweeps. The bogatyr wants to reduce dragon's firepower as much as possible. What is the maximum total firepower of heads which Ilya can cut with at most two sword sweeps?

Input

The first line contains a pair of integer numbers?n?and?k?(1?≤?n,?k?≤?2·10⁵) — the number of Gorynych's heads and the maximum number of heads Ilya can cut with a single sword sweep. The second line contains the sequence of integer numbers?f₁,?f₂,?...,?f_n?(1?≤?f_i?≤?2000), where?f_i?is the firepower of the?i-th head.

Output

Print the required maximum total head firepower that Ilya can cut.

Examples

Input 8 2
1 3 3 1 2 3 11 1 Output 20 Input 4 100
10 20 30 40 Output100 題意：
　　給你兩個(gè)數(shù)n和k，然后n個(gè)數(shù)，進(jìn)行兩次操作，一次最多可以消除連續(xù)的K個(gè)數(shù)，然后剩余的數(shù)又變成連續(xù)的，問兩次操作后，消除的數(shù)的總和最大為多少。
題解：
　　dp[i][1]表示的是1-i的區(qū)間中，第一次操作能得到的最大值，因?yàn)閕從k-n,每次加一，所以每次對(duì)最大值可能造成改變的只有新加進(jìn)來的數(shù)，dp[i][1]=max(dp[i-1][1],sum[i]-sum[i-k]);
　　dp[i][1]表示的是1-i的區(qū)間中，兩次操作后能得到的最大值，dp[i][2]=max(dp[i-1][2],sum[i]-sum[i-k]+dp[i-k][1]);對(duì)于兩次操作，狀態(tài)轉(zhuǎn)移方程就是1- i-1區(qū)間兩次操作后最大值或者是新加入的i影響的區(qū)間sum[i]-sum[i-k]+1- i-k區(qū)間操作一次的最大值。 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int maxn=2e5+10; 7 int sum[maxn],a[maxn],dp[maxn][3]; 8 int main() 9 { 10 int n,k; 11 scanf("%d%d",&n,&k); 12 for(int i=1;i<=n;i++) 13 { 14 scanf("%d",&a[i]); 15 sum[i]=sum[i-1]+a[i]; 16 } 17 if(2*k>=n) 18 { 19 printf("%d\n",sum[n]); 20 return 0; 21 } 22 memset(dp,0,sizeof(dp)); 23 for(int i=k;i<=n;i++) 24 { 25 dp[i][1]=max(dp[i-1][1],sum[i]-sum[i-k]); 26 dp[i][2]=max(dp[i-1][2],sum[i]-sum[i-k]+dp[i-k][1]); 27 // printf("%d %d %d\n",i,dp[i][1],dp[i][2]); 28 } 29 printf("%d\n",dp[n][2]); 30 return 0; 31 }

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轉(zhuǎn)載于:https://www.cnblogs.com/1013star/p/10070083.html

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