TOJ 4244: Sum

4244: Sum  

Time Limit(Common/Java):3000MS/9000MS     Memory Limit:65536KByte
Total Submit: 63            Accepted:10

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.

Input

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 10⁸ indicating an element of the sequence.

Output

Output the requested sum.

Sample Input

4
3
1
7
2

Sample Output

31

Hint

The consecutive subsequence of the sequence (3, 1, 7, 2) has:
(3, 1), (1, 7), (7, 2), (3, 1, 7), (1, 7, 2), (3, 1, 7, 2)

The sum of all values equals 2+6+5+6+6+6=31

这题看起来很简单啊，然后自然而然就想到了朴素的O(n^2)做法，然而TLE，仔细一想要不用dp做，保存当前最大值，可是不还是O(n^2)，虽然循环次数少点，堆排序TLE同理。那天坐CF做完了无聊了就又在想这道题，想不出来就在群里问了问，大神给了我单调栈的做法和sort排序找位置的方法

sort大法来自010大神 点我获得代码

我的单调栈就是向左向右找到其可以左延伸右延伸的位置，排列组合就好了（左边包括他自己有n个，右侧包括他自己有m个，他的贡献就是（mn-1）个）

然后最大值来一次，最小值来一次，over

和普通的单调栈一样，这个的话就是检查扩展，可以扩展就去和前面那个值一个位置，其实求的就是最多扩展到哪里。一个值最多被访问两次，和单调栈一样都是2*n

单调栈也是，访问这个值，进队，可能还要出队。虽然是for套while但是复杂度还是n

#include <cstdio>
const int N=;
__int64 a[N];
int L[N],R[N];
int main() {
    int n;
    while(~scanf("%d",&n)) {
        for(int i=; i<=n; i++) {
            scanf("%I64d",&a[i]);
            L[i]=i;
            R[i]=i;
        }
        a[]=-;
        a[n+]=-;
        for(int i = ; i <= n; i++) {
            while(a[i] <= a[L[i] - ])
                L[i] = L[L[i] - ];
        }
        for(int i = n; i >= ; i--) {
            while(a[i]< a[R[i] + ])
                R[i] = R[R[i] + ];
        }
        __int64 sum=;
        for(int i = ; i <= n; i++) {
            sum-=a[i]*(i-L[i]+)*(R[i]-i+);
        }
        a[]=<<;
        a[n+]=<<;
        for(int i=; i<=n; i++) {
            L[i]=i;
            R[i]=i;
        }
        for(int i = ; i <= n; i++) {
            while(a[i] >= a[L[i] - ])
                L[i] = L[L[i] - ];
        }
        for(int i = n; i >= ; i--) {
            while(a[i] > a[R[i] + ])
                R[i] = R[R[i] + ];
        }
        for(int i = ; i <= n; i++) {
            sum+=a[i]*(i-L[i]+)*(R[i]-i+);
        }
        printf("%I64d\n",sum);
    }
    return ;
}