TOJ4244: Sum

description

Given a sequence, we define the seqence's value equals the difference between the largest element and the smallest element in the sequence. As an example, the value of sequence (3, 1, 7, 2) = 7-1 = 6.

Now, given a sequence S, output the sum of all value of consecutive subsequences.

Entry

The first line has an integer N (2 ≤ N ≤ 300000) indicating the number of elements of the sequence. Then follows N lines, each line has a positive integer no larger than 10⁸indicating an element of the sequence.

Export

Output the requested sum.

Sample input

4
3
1
7
2

Sample Output

Problem-solving ideas: seeking simplification formula available to all sections and maximum - minimum and all sections. Maintenance monotonous look, calculate the contribution of each element.

#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int N=3e5+5;
LL a[N],st[N];
int main()
{
    int n,m,k;
    cin>>n;
    for(int i=1;i<=n;i++) cin>>a[i];    
    LL ans=0,sum=0,top=0;    
    for(int i=1;i<=n;i++){
        while(top>0&&a[i]>a[st[top]]){
            LL num=a[st[top]];
            sum-=num*(st[top]-st[top-1]);
            top--; 
        }
        st[++top]=i;
        sum+=a[i]*(st[top]-st[top-1]);
        ans+=sum;
    }//区间最大值和 
    sum=0,top=0;
    for(int i=1;i<=n;i++){
        while(top>0&&a[i]<a[st[top]]){
            NUM LL=A [ST [Top]]; 
            SUM - NUM = * (ST [Top] -st [TOP- . 1 ]); 
            Top - ; 
        } 
        ST [ ++ Top] = I; 
        SUM + A = [I] * ( ST [Top] -st [TOP- . 1 ]); 
        ANS - = SUM; 
    } // interval and a minimum value 
    COUT ANS << << endl; 
}

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