HDU 5805 NanoApe Loves Sequence
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Problem Description
NanoApe, the Retired Dog, has returned back to prepare for the National Higher Education Entrance Examination!
In math class, NanoApe picked up sequences once again. He wrote down a sequence withn numbers on the paper and then randomly deleted a number in the sequence. After that, he calculated the maximum absolute value of the difference of each two adjacent remained numbers, denoted as F .
Now he wants to know the expected value ofF , if he deleted each number with equal probability.
In math class, NanoApe picked up sequences once again. He wrote down a sequence with
Now he wants to know the expected value of
Input
The first line of the input contains an integer T , denoting the number of test cases.
In each test case, the first line of the input contains an integern , denoting the length of the original sequence.
The second line of the input containsn integers A1,A2,...,An , denoting the elements of the sequence.
1≤T≤10, 3≤n≤100000, 1≤Ai≤109
In each test case, the first line of the input contains an integer
The second line of the input contains
Output
For each test case, print a line with one integer, denoting the answer.
In order to prevent using float number, you should print the answer multiplied byn .
In order to prevent using float number, you should print the answer multiplied by
Sample Input
141 2 3 4
Sample Output
6
可以用二分,dp
主要思想还是dp
#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#include<queue>#include<stack>using namespace std;#define ll long longconst int N=100010;int t,n,a[N],d[N],l[N],r[N];int main(){ int i,j; scanf("%d",&t); while (t--) { scanf("%d",&n); for (i=1;i<=n;i++) scanf("%d",&a[i]); for (i=1;i<n;i++) d[i]=abs(a[i]-a[i+1]); l[0]=0; for (i=1;i<n;i++) l[i]=d[i]>l[i-1]?d[i]:l[i-1]; r[n]=0; for (i=n-1;i>0;i--) r[i]=d[i]>r[i+1]?d[i]:r[i+1]; ll ans=0; int tmp; for (i=1;i<=n;i++) { if (i==1) ans+=r[2]; else if (i==n) ans+=l[n-2]; else { tmp=abs(a[i-1]-a[i+1]); if (l[i-2]>tmp) tmp=l[i-2]; if (r[i+1]>tmp) tmp=r[i+1]; ans+=tmp; } } printf("%lld\n",ans); } return 0;}
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