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
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.
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
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
",num-minn);
31 }
32 }
33 return 0;
34 }