Alice and Bob are playing yet another card game. This time the rules are the following. There are nn cards lying in a row in front of them. The ii-th card has value aiai.
First, Alice chooses a non-empty consecutive segment of cards [l;r][l;r] (l≤rl≤r). After that Bob removes a single card jj from that segment (l≤j≤r)(l≤j≤r). The score of the game is the total value of the remaining cards on the segment (al+al+1+⋯+aj−1+aj+1+⋯+ar−1+ar)(al+al+1+⋯+aj−1+aj+1+⋯+ar−1+ar). In particular, if Alice chooses a segment with just one element, then the score after Bob removes the only card is 00.
Alice wants to make the score as big as possible. Bob takes such a card that the score is as small as possible.
What segment should Alice choose so that the score is maximum possible? Output the maximum score.Input
The first line contains a single integer nn (1≤n≤1051≤n≤105) — the number of cards.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (−30≤ai≤30−30≤ai≤30) — the values on the cards.Output
Print a single integer — the final score of the game.ExamplesinputCopy
5
5 -2 10 -1 4
outputCopy
6
inputCopy
8
5 2 5 3 -30 -30 6 9
outputCopy
10
inputCopy
3
-10 6 -15
outputCopy
0
Note
In the first example Alice chooses a segment [1;5][1;5] — the entire row of cards. Bob removes card 33 with the value 1010 from the segment. Thus, the final score is 5+(−2)+(−1)+4=65+(−2)+(−1)+4=6.
In the second example Alice chooses a segment [1;4][1;4], so that Bob removes either card 11 or 33 with the value 55, making the answer 5+2+3=105+2+3=10.
In the third example Alice can choose any of the segments of length 11: [1;1][1;1], [2;2][2;2] or [3;3][3;3]. Bob removes the only card, so the score is 00. If Alice chooses some other segment then the answer will be less than 00.
题意： 要求找到“区间和-区间最大值”的最大值
思路：如果通过枚举区间和去思考十分难思考。
思维转化一下：我们通过枚举区间最大值求思考。
对一段区间，如果全都是负数，那么区间和 -区间最大值的最大值一定是0；
我们枚举区间最大值，当累加的连续子段和的下一个a[j]>i的时候，我们把sum重置=0；并且开始新的sum累加。在sum<0的情况下，我们也重置sum=0（因为sum最小为0并且这段sum<0对以后的答案没有贡献。可以回想最大连续子段和的累加）；然后我们不断更新最大值即可
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
typedef long long LL;
LL a[maxn];
int main(void)
{
LL n;cin>>n;
for(LL i=1;i<=n;i++) cin>>a[i];
//枚举区间最大值
LL ans=-0x3f3f3f3f;
//区间最大值要从非负数开始枚举。
//因为如果区间最大值是负数，那么最大子段和的必然是0（只取一个）
//不会有下面的sum的变化，可以用特判解决
for(LL i=0;i<=30;i++)
{
LL sum=0;//全是负数最差的情况sum=0;因为要减去唯一的负数
//枚举连续最大子段和
for(LL j=1;j<=n;j++)
{
if(a[j]>i)//不符合区间内<=区间最大值的假设
{
sum=0;continue;
}
sum+=a[j];
if(sum<0) sum=0;
ans=max(ans,sum-i);
}
}
cout<<ans<<endl;
return 0;
}