我是
靠谱客的博主
危机香菇,这篇文章主要介绍
stars,现在分享给大家,希望可以做个参考。
B - Stars
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit
Status
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.
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.
#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
const int maxn = 32005;
int c[maxn];
int ans[maxn];
int lowbit(int x) {
return x&(-x);
}
void add(int x){
while(x<=maxn){
c[x]+=1; x+=lowbit(x);
}
}
int sum(int x){
int ret=0;
while(x>0){
ret+=c[x]; x-=lowbit(x);
}
return ret;
}
int main(){
int n,x,y,i;
while(scanf("%d",&n)!=EOF){
memset(ans,0,sizeof(ans));
memset(c,0,sizeof(c));
for(i=0;i<n;i++){
scanf("%d %d",&x,&y);
ans[ sum(++x)]++;
add(x);
}
for(i=0;i<n;i++)
printf("%dn",ans[i]);
}
return 0;
}
最后
以上就是危机香菇最近收集整理的关于stars的全部内容,更多相关stars内容请搜索靠谱客的其他文章。
本图文内容来源于网友提供,作为学习参考使用,或来自网络收集整理,版权属于原作者所有。
发表评论 取消回复