【CF Edu 28 C. Four Segments】

time limit per test 1 second

memory limit per test 256 megabytes

input standard input

output standard output

You are given an array of n integer numbers. Let sum(l, r) be the sum of all numbers on positions from l to r non-inclusive (l-th element is counted, r-th element is not counted). For indices l and r holds 0 ≤ l ≤ r ≤ n. Indices in array are numbered from 0.

For example, if a = [ - 5, 3, 9, 4], then sum(0, 1) =  - 5, sum(0, 2) =  - 2, sum(1, 4) = 16 and sum(i, i) = 0 for each i from 0 to 4.

Choose the indices of three delimiters delim₀, delim₁, delim₂ (0 ≤ delim₀ ≤ delim₁ ≤ delim₂ ≤ n) and divide the array in such a way that the value of res = sum(0, delim₀) - sum(delim₀, delim₁) + sum(delim₁, delim₂) - sum(delim₂, n) is maximal.

Note that some of the expressions sum(l, r) can correspond to empty segments (if l = r for some segment).

Input

The first line contains one integer number n (1 ≤ n ≤ 5000).

The second line contains n numbers a₀, a₁, ..., a_n - 1 ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Choose three indices so that the value of res is maximal. If there are multiple answers, print any of them.

Examples

input

3
-1 2 3

output

0 1 3

input

4
0 0 -1 0

output

0 0 0

input

1
10000

output

1 1 1

【翻译】给出一个长度为n的序列(1~n)，现在需要选择三个点d1,d2,d2(0<=d1<=d2<=d3<=n),设ABCD分别为区间(0,d1],(d1,d2],(d2,d3],(d3,n]的内部元素和,求A-B+C-D取到最大值时d1,d2,d3的值，情况多种就随便输出一种。

题解：
    ①好像怎么弄都是O(n2),枚举d2,然后取d1,d3的最有位置更新答案。

    ②为辅助上述方法,预处理(A-B),(C-D)的最大值，并记录取最大值时候d的位置。

    ③预处理会使用到前缀和。

#include<stdio.h>
#define ll long long
#define comb (val1[i]+val2[i])
#define S(l,r) (sum[r]-sum[l-1])
#define go(i,a,b) for(int i=a;i<=b;i++)
const int N=5003;int n,p1[N],p3[N],P1,P2,P3;
ll a[N],_[N],sum[N],val1[N],val2[N],ans,t;
int main()
{	
	scanf("%d",&n);ans=1ll*-1e9*1e9;
	
	go(i,1,n)scanf("%I64d",a+i),sum[i]=sum[i-1]+a[i];
	go(i,0,n){val1[i]=1ll*-1e9*1e9;
	go(j,0,i)if((t=S(1,j)-S(j+1,i))>val1[i])p1[i]=j,val1[i]=t;}
	go(i,0,n){val2[i]=1ll*-1e9*1e9;
	go(j,i,n)if((t=S(i+1,j)-S(j+1,n))>val2[i])p3[i]=j,val2[i]=t;}
	go(i,0,n)if(val1[i]+val2[i]>ans)ans=comb,P1=p1[i],P2=i,P3=p3[i];
		
	printf("%d %d %d
",P1,P2,P3);return 0;
}//Paul_Guderian

挥挥手倦鸟飞过丛林，隐没在碎与溃的深谷。——————汪峰《挥挥手》