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 10⁸indicating an element of the sequence.

Output

Output the requested sum.

Sample Input

4
3
1
7
2

Sample Output

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 ;

}

巴特西

TOJ 4244: Sum

4244: Sum

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