HDU 6227.Rabbits-规律 (2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）)

 

Rabbits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1193    Accepted Submission(s): 628

Problem Description

Here N (N ≥ 3) rabbits are playing by the river. They are playing on a number line, each occupying a different integer. In a single move, one of the outer rabbits jumps into a space between any other two. At no point may two rabbits occupy the same position.
Help them play as long as possible

 

 

Input

The input has several test cases. The first line of input contains an integer t (1 ≤ t ≤ 500) indicating the number of test cases.
For each case the first line contains the integer N (3 ≤ N ≤ 500) described as above. The second line contains n integers a1 < a2 < a3 < ... < aN which are the initial positions of the rabbits. For each rabbit, its initial position
ai satisfies 1 ≤ ai ≤ 10000.

 

 

Output

For each case, output the largest number of moves the rabbits can make.

 

 

Sample Input

5

3

3 4 6

3

2 3 5

3

3 5 9

4

1 2 3 4

4

1 2 4 5

 

 

Sample Output

1

1

3

0

1

 

 

Source

2017ACM/ICPC亚洲区沈阳站-重现赛（感谢东北大学）

 

 

 

 

代码:

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e5+10;
 4 const int INF=0x3f3f3f3f;
 5 int flag[N];
 6 int a[N];
 7 int main(){
 8     int t,n,minn;
 9     int num1,num2,num;
10     while(~scanf("%d",&t)){
11         while(t--){
12            scanf("%d",&n);
13            memset(flag,0,sizeof(flag));
14            for(int i=0;i<n;i++){
15                 scanf("%d",&a[i]);
16                 flag[a[i]]=1;
17            }
18            num1=0,num2=0,num=0;
19            for(int i=a[0];i<=a[1];i++){
20                 if(flag[i]==0)num1++;
21            }
22            for(int i=a[n-2];i<=a[n-1];i++){
23                 if(flag[i]==0)num2++;
24            }
25            for(int i=a[0];i<=a[n-1];i++){
26                 if(flag[i]==0)num++;
27            }
28            //cout<<num1<<" "<<num2<<" "<<num<<endl;
29            minn=min(num1,num2);
30            printf("%d\n",num-minn);
31         }
32     }
33     return 0;
34 }