Equal Cut

Posted 2021-01-25 hao-tian

Snuke has an integer sequence A of length N.

He will make three cuts in A and divide it into four (non-empty) contiguous subsequences B,C,D and E. The positions of the cuts can be freely chosen.

Let P,Q,R,S be the sums of the elements in B,C,D,E, respectively. Snuke is happier when the absolute difference of the maximum and the minimum among P,Q,R,S is smaller. Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

Constraints
4≤N≤2×105
1≤Ai≤109
All values in input are integers.

 

输入

Input is given from Standard Input in the following format:

N
A1 A2 … AN

 

输出

Find the minimum possible absolute difference of the maximum and the minimum among P,Q,R,S.

 

样例输入

复制样例数据

5
3 2 4 1 2

样例输出

2

 

提示

If we divide A as B,C,D,E=(3),(2),(4),(1,2), then P=3,Q=2,R=4,S=1+2=3. Here, the maximum and the minimum among P,Q,R,S are 4 and 2, with the absolute difference of 2. We cannot make the absolute difference of the maximum and the minimum less than 2, so the answer is 2.

 

//枚举中点位置 再根据中点位置 贪心l，r的位置 代码如下 参考：https://blog.csdn.net/aaakirito/article/details/80884168?utm_source=blogxgwz5

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
const int maxn = 200005;
const ll inf = 0x7fffffff;
ll a[maxn];
ll sum[maxn];
int main()
{
   // cout<<inf<<endl;
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        a[i]=read();
        sum[i]+=sum[i-1]+a[i];
    }
    int l=1,r=3;
    ll minn = inf;
    for(int i=2;i<n-1;i++)
    {
        while(l<i&&abs((sum[i]-sum[l])-(sum[l]-sum[0]))>=abs((sum[i]-sum[l+1])-(sum[l+1]-sum[0])))
        {
            l++;
        }
        while(r<n&&abs((sum[r]-sum[i])-(sum[n]-sum[r]))>abs((sum[r+1]-sum[i])-(sum[n]-sum[r+1])))
        {
            r++;
        }
        ll x,y,p,q;
        x=sum[i]-sum[l];
        y=sum[l]-sum[0];
        p=sum[r]-sum[i];
        q=sum[n]-sum[r];
        minn = min(minn,max(x,max(p,max(y,q)))-min(x,min(y,min(p,q))));
    }
    printf("%lld
",minn);
}