P is a permutation of the integers from 1 to N(index starting from 1).
Here is the code of Bubble Sort in C++.

for(int i=1;i<=N;++i)
    for(int j=N,t;j>i;—j)
        if(P[j-1] > P[j])
            t=P[j],P[j]=P[j-1],P[j-1]=t;

After the sort, the array is in increasing order. ?? wants to know the absolute values of difference of rightmost place and leftmost place for every number it reached.InputThe first line of the input gives the number of test cases T; T test cases follow.
Each consists of one line with one integer N, followed by another line with a permutation of the integers from 1 to N, inclusive.

limits
T <= 20
1 <= N <= 100000
N is larger than 10000 in only one case.
OutputFor each test case output “Case #x: y1 y2 … yN” (without quotes), where x is the test case number (starting from 1), and yi is the difference of rightmost place and leftmost place of number i.Sample Input

2
3
3 1 2
3
1 2 3

Sample Output

Case #1: 1 1 2
Case #2: 0 0 0


        
 

Hint

In first case, (3, 1, 2) -> (3, 1, 2) -> (1, 3, 2) -> (1, 2, 3)
the leftmost place and rightmost place of 1 is 1 and 2, 2 is 2 and 3, 3 is 1 and 3
In second case, the array has already in increasing order. So the answer of every number is 0.

        
 求一個數列在冒泡排序中每個數字所能到達的最左和最右位置的距離，最左即為min（目標位置，當前位置），最右是後面有多少個比它小的數+當前位置。
簡單模擬即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 
 7 using namespace std;
 8 
 9 const int maxn=100005;
10 int c[maxn],d[maxn],g[maxn],h[maxn];
11 
12 void update(int x,int v)
13 {
14     for(;x<=maxn;x+=x&(-x))
 
15         g[x]+=v;
16 }
17 
18 int getsum(int x)
19 {
20     int sum=0;
21     for(;x>0;x-=x&(-x))
22         sum+=g[x];
23     return sum;
24 }
25 
26 int main()
27 {
28     int n,T,s=1;
29     scanf("%d",&T);
30     while(T--)
31     {
32         scanf("%d",&n);
33         memset(c,0,sizeof(c));
34         memset(d,0,sizeof(d));
35         memset(g,0,sizeof(g));
36         for(int i=1;i<=n;i++)
37             scanf("%d",&c[i]);
38         for(int i=1;i<=n;i++)
39         {
40             d[c[i]]=min(c[i],i);
41         }
42         for(int i=n;i>=1;i--)
43         {
44             h[c[i]]=getsum(c[i])+i;
45             update(c[i],1);
46         }
47         printf("Case #%d: %d",s++,abs(d[1]-h[1]));
48         for(int i=2;i<=n;i++)
49         {
50             printf("% d",abs(d[i]-h[i]));
51         }
52         cout<<endl;
53     }
54     
55     
56     return 0;    
57 } 

HDU - 5775 Bubble Sort