TOJ 4244: Sum
4244: Sum
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=300005; __int64 a[N]; int L[N],R[N]; int main() { int n; while(~scanf("%d",&n)) { for(int i=1; i<=n; i++) { scanf("%I64d",&a[i]); L[i]=i; R[i]=i; } a[0]=-1; a[n+1]=-1; for(int i = 1; i <= n; i++) { while(a[i] <= a[L[i] - 1]) L[i] = L[L[i] - 1]; } for(int i = n; i >= 1; i--) { while(a[i]< a[R[i] + 1]) R[i] = R[R[i] + 1]; } __int64 sum=0; for(int i = 1; i <= n; i++) { sum-=a[i]*(i-L[i]+1)*(R[i]-i+1); } a[0]=1<<30; a[n+1]=1<<30; for(int i=1; i<=n; i++) { L[i]=i; R[i]=i; } for(int i = 1; i <= n; i++) { while(a[i] >= a[L[i] - 1]) L[i] = L[L[i] - 1]; } for(int i = n; i >= 1; i--) { while(a[i] > a[R[i] + 1]) R[i] = R[R[i] + 1]; } for(int i = 1; i <= n; i++) { sum+=a[i]*(i-L[i]+1)*(R[i]-i+1); } printf("%I64d\n",sum); } return 0; }
本文来自博客园,作者:暴力都不会的蒟蒻,转载请注明原文链接:https://www.cnblogs.com/BobHuang/p/6832480.html