http://acm.hdu.edu.cn/showproblem.php?pid=5775

Bubble Sort

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.

题意：求出1-n里面各个数在冒泡排序里面可以达到的最右距离和最左距离之差是多少。

 1 #include <cstring>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <iostream>
 5 #include <cmath>
 6 using namespace std;
 7 #define N 100005
 8 /*
 9 树状数组
10 从右边往左扫看右边有多少个数是小于当前位置的数的，
11 然后可以达到的最右距离就是最初始的位置加上小于它的数的数量
12 可以达到的最左距离是min(最初始的位置,i)
13 所以最右减去最左就是i的答案了.
14 */
15 int bit[N];
16 int num[N], tmp[N], r[N];
17
18 int lowbit(int x)
19 {
20     return x & (-x);
21 }
22
23 void update(int x)
24 {
25     while(x <= N) {
26         bit[x] += 1;
27         x += lowbit(x);
28     }
29 }
30
31 int query(int x)
32 {
33     int ans = 0;
34     while(x) {
35         ans += bit[x];
36         x -= lowbit(x);
37     }
38     return ans;
39 }
40
41 int main()
42 {
43     int t;
44     scanf("%d", &t);
45     for(int cas = 1; cas <= t; cas++) {
46         memset(bit, 0, sizeof(bit));
47         int n;
48         scanf("%d", &n);
49         for(int i = 1; i <= n; i++){
50             scanf("%d", num+i);
51             tmp[num[i]] = i;
52         }
53         for(int i = n; i > 0; i--) {
54             update(num[i]);
55             r[num[i]] = query(num[i] - 1);
56         }
57
58         printf("Case #%d: ", cas);
59         for(int i = 1; i <= n; i++) {
60             if(i != n) printf("%d ", abs(tmp[i]+r[i] - min(tmp[i], i)));
61             else printf("%d\n", abs(tmp[i]+r[i] - min(tmp[i], i)));
62         }
63     }
64     return 0;
65 }