标签：output 外挂 NPU names its ane class limit The
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
Hint
This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.
1 //#include <bits/stdc++.h>
2 #include <cstdio>
3 using namespace std;
4 #define lowbit(x) x&(-x)
5 const int MAXN=32005;
6 int a[MAXN],leve[MAXN];
7 int n;
8 //int lowbit(int x)
9 //{
10 // return x&(-x);
11 //}
12 int sum(int x)
13 {
14 int ans=0;
15 while(x)
16 {
17 ans+=a[x];
18 x-=lowbit(x);
19 }
20 return ans;
21 }
22 void add(int x,int v)
23 {
24 while (x<=MAXN)
25 {
26 a[x]+=v;
27 x+=lowbit(x);
28 }
29
30 }
31 template <class T>
32 inline void scan_d(T &ret)
33 {
34 char c;
35 ret = 0;
36 while ((c = getchar()) < ‘0‘ || c > ‘9‘);
37 while (c >= ‘0‘ && c <= ‘9‘)
38 {
39 ret = ret * 10 + (c - ‘0‘), c = getchar();
40 }
41 }
42 inline Out(int a)
43 { // 输出外挂
44 if (a < 0)
45 {
46 putchar(‘-‘);
47 a = -a;
48 }
49 if (a >= 10)
50 {
51 Out(a / 10);
52 }
53 putchar(a % 10 + ‘0‘);
54 }
55 int main()
56 {
57 scanf("%d",&n);
58 int x,y;
59 for (int i = 0; i <n ; ++i) {
60 scan_d(x);
61 scan_d(y);
62 x++; //切记此处加1，因为x是从0开始的，如果没有添加会造成超时，因为0+lowbit(0)=0;死循环
63 leve[sum(x)]++;
64 add(x,1);
65 }
66 for (int i = 0; i <n ; ++i) {
67 Out(leve[i]);
68 putchar(‘\n‘);
69 }
70 return 0;
71 }
POJ2352 Stars
标签：output 外挂 NPU names its ane class limit The
原文地址：https://www.cnblogs.com/-xiangyang/p/9404135.html
