终于在cf写出了一道正经的dp题，值得庆贺(￣▽￣)~*
dp[i]代表以i结尾的长度为偶数的贡献最大的反转区间，贡献是奇数位的和与偶数位的差
然后转移方程：奇数就是dp[i]=max(dp[i],dp[i-2]+a[i]-a[i-1]);偶数dp[i]=max(dp[i],dp[i-2]+a[i-1]-a[i]);
反转区间长度必为偶数，因为如果是奇数那么相当于没反转，那这么dp为什么能保证都选的是偶数长度的区间呢？dp[0]不选，dp[1]=max(0,dp[1]-dp[0])即dp[1]只能选0个或两个，之后都是两个数两个数往前面dp[i-2]连或不连,所以必为偶
答案就是原序列偶数位和加上dp[i]中最大值
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a0, the second one is a1
, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a
with borders l and r is a[l;r]=al,al+1,…,ar
.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a0,a2,…,a2k
for integer k=⌊n−12⌋
should be maximum possible).
You have to answer t
independent test cases.
Input
The first line of the input contains one integer t
(1≤t≤2⋅104) — the number of test cases. Then t
test cases follow.
The first line of the test case contains one integer n
(1≤n≤2⋅105) — the length of a. The second line of the test case contains n integers a0,a1,…,an−1 (1≤ai≤109), where ai is the i-th element of a
.
It is guaranteed that the sum of n
does not exceed 2⋅105 (∑n≤2⋅105
).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a
.
Example
Input
Copy
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
Copy
26
5
37
5
#include <cstdio>
#include<iostream>
#include <cstring>
#include <algorithm>
#include <stack>
#include<set>
#include<vector>
using namespace std;
const int maxn=1505000;
#define mod 1000000007
#include<stdio.h>
#define ll long long
ll dp[maxn],a[maxn];
int main()
{
ll t;
scanf("%lld",&t);
while(t--)
{
ll n;
scanf("%lld",&n);
for(ll i=0;i<n;i++) scanf("%lld",&a[i]),dp[i]=0;
if(n==1)
{
printf("%lld\n",a[0]);
continue;
}
dp[0]=0;dp[1]=max(dp[1],a[1]-a[0]);
ll mx=max(0ll,dp[1]);
for(ll i=2;i<n;i++)
{
if(i%2==0)
dp[i]=max(dp[i],dp[i-2]+a[i-1]-a[i]);
else
dp[i]=max(dp[i],dp[i-2]+a[i]-a[i-1]);
//printf("dp[%lld]=%lld\n",i,dp[i]);
mx=max(mx,dp[i]);
}
ll ans=0;
for(ll i=0;i<n;i+=2) ans+=a[i];
printf("%lld\n",ans+mx);
}
}