Description
Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 
For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it‘s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3.
You are to write a program that will count the amounts of the stars of each level on a given map.
Input
The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate.
Output
The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.
Sample Input
5 1 1 5 1 7 1 3 3 5 5
Sample Output
1 2 1 1 0题意：给出一个星星的坐标，问从（0,0）到（x,y）的矩形区域内包含了不含有本身的星星的个数，然后从0-->(n-1)输出思路：首先由于y是从小到大输入的所以只要对x进行统计就行了。介绍一下树状数组：线段树通过对x->min(2^n)以及以上2^n进行更新就表示在其后方插入了一个数值即为（Up函数）当统计时就是计算其前方和最近的2^n的和。给出一个树状数组的链接：http://blog.csdn.net/int64ago/article/details/7429868AC代码：
1 #include <iostream>
2 #include <algorithm>
3 #include <cstdio>
4 #include <cstring>
5 using namespace std;
6 int a[40000],b[40000],c[40000];
7 int lowbit(int x)
8 {
9 return x&(-x);
10 }
11 //增加一个元素
12 int up(int x)
13 {
14 while(x<35000)
15 {
16 a[x]++;
17 x=x+lowbit(x);
18 }
19 return 0;
20 }
21 //求和
22 int sum(int x)
23 {
24 int ans;
25 ans=0;
26 while(x)
27 {
28 ans=ans+a[x];
29 x=x-lowbit(x);
30 }
31 return ans;
32 }
33 int main()
34 {
35 int n,k,flag;
36 int x,y;
37 while(~scanf("%d\n",&n))
38 {
39 k=0;
40 memset(a,0,sizeof(a));
41 memset(b,0,sizeof(b));
42 //memset(c,0,sizeof(c));
43 for(int i=0; i<n; i++)
44 {
45 scanf("%d%d",&x,&y);
46 x++;
47 b[sum(x)]++;
48 up(x);
49 }
50 for(int i=0; i<n; i++)
51 {
52 printf("%d\n",b[i]);
53 }
54 }
55 return 0;
56 }