# Bubble Sort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 591    Accepted Submission(s): 359

Problem Description

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.

Input

The 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.

Output

For 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.

```#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int C[maxn];
int n;

inline int lowbit(int x) {
return x & (-x);
}

int sum(int x) {
int ret = 0;
while(x > 0) {
ret += C[x]; x -= lowbit(x);
}
return ret;
}

void add(int x, int d) {
while(x <= maxn) {
C[x] += d; x += lowbit(x);
}
}

int a[maxn];
int r[maxn], l[maxn];
int id[maxn];

int main() {
int t;
scanf("%d", &t);
int cas = 0;
while(t--) {
memset(C, 0, sizeof(C));
scanf("%d", &n);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);

for(int i = n; i >= 1; i--) {
r[i] = sum(a[i] - 1);
id[a[i]] = i;
}

//for(int i = 1; i <= n; i++) printf("%d ", r[i]); puts("");
vector<int> ans;
for(int i = 1; i <= n; i++) {
int ll = min(id[i], i);
int rr = id[i] + r[id[i]];
//printf("check %d %d %d\n", i, ll, rr);
ans.push_back(rr - ll);
}
printf("Case #%d:", ++cas);
for(int i = 0; i < ans.size(); i++) {
printf(" %d", ans[i]);
}puts("");
}
}```

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