NanoApe Loves Sequence-待解决
NanoApe Loves Sequence
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 440 Accepted Submission(s): 205
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 with n![]()
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 of F![]()
, if he deleted each number with equal probability.
In math class, NanoApe picked up sequences once again. He wrote down a sequence with n
Now he wants to know the expected value of F
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 integer n![]()
, denoting the length of the original sequence.
The second line of the input contains n![]()
integers A![]()
1![]()
,A![]()
2![]()
,...,A![]()
n![]()
![]()
, denoting the elements of the sequence.
1≤T≤10, 3≤n≤100000, 1≤A![]()
i![]()
≤10![]()
9![]()
![]()
In each test case, the first line of the input contains an integer n
The second line of the input contains n
1≤T≤10, 3≤n≤100000, 1≤A
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 by n![]()
.
In order to prevent using float number, you should print the answer multiplied by n
Sample Input
1 4 1 2 3 4
Sample Output
6
Source
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#include <iostream> #include <cstdio> #include <cmath> using namespace std; int main() { int t; int n; int cha=0; int cha2=0; int a[100005]={0}; int maxx1; int first; int maxx2; int second; int maxx3; int third; int sum=0; scanf("%d",&t); for(int z=0;z<t;z++){ sum=0; maxx1=0; maxx2=0; maxx3=0; scanf("%d",&n); for(int i=0;i<n;i++){ scanf("%d",&a[i]); if(i!=0){ cha=abs(a[i]-a[i-1]); if(maxx1<cha){ maxx1=cha; first=i; } } } for(int i=1;i<n;i++){ cha=abs(a[i]-a[i-1]); if(maxx2<cha){ if(maxx2<=maxx1){ if(i==first){ continue; }else{ maxx2=cha; second=i; } } } } for(int i=1;i<n;i++){ cha=abs(a[i]-a[i-1]); if(maxx2<cha){ if(maxx3<=maxx1&&maxx3<=maxx2){ if(i==first||i==second){ continue; }else{ maxx3=cha; third=i; } } } } for(int i=1;i<n-1;i++){ cha2=abs(a[i+1]-a[i-1]); if(maxx1<=cha2){ sum+=cha2; } if(maxx1>cha2){ if(i==first&&i+1==second||i==second&&i+1==first){ if(maxx3!=0){ if(maxx3>=cha2){ sum+=maxx3; }else{ sum+=cha2; } }else{ sum+=cha2; } } if(i==first||i==first-1){ if(cha2<=maxx2){ sum+=maxx2; }else{ sum+=cha2; } }else{ sum+=maxx1; } } } if(first==1){ sum+=maxx2; sum+=maxx1; } if(first==n-1){ sum+=maxx2; sum+=maxx1; } if(first!=1&&first!=n-1){ sum+=(2*maxx1); } printf("%d\n",sum); } return 0; }
