|
招直播！看直播，投简历....
|
Color the ballTime Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 25201 Accepted Submission(s): 12246
Problem Description
N个气球排成一排，从左到右依次编号为1,2,3....N.每次给定2个整数a b(a <= b),lele便为骑上他的“小飞鸽"牌电动车从气球a开始到气球b依次给每个气球涂一次颜色。但是N次以后lele已经忘记了第I个气球已经涂过几次颜色了，你能帮他算出每个气球被涂过几次颜色吗？
Input
每个测试实例第一行为一个整数N,(N <= 100000).接下来的N行，每行包括2个整数a b(1 <= a <= b <= N)。
当N = 0，输入结束。
Output
每个测试实例输出一行，包括N个整数，第I个数代表第I个气球总共被涂色的次数。
Sample Input
3 1 1 2 2 3 3 3 1 1 1 2 1 3 0
Sample Output
1 1 1 3 2 1
Author
8600
Source
HDU 2006-12 Programming Contest
Recommend
LL | We have carefully selected several similar problems for you: 1542 1394 1698 1255 2795
|
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int n;
int num[maxn];
int lowbit(int x)
{return x&(-x);
}
void update(int pos,int surplus)
{while(pos<=n){num[pos]+=surplus;pos+=lowbit(pos); }
}
int getsum(int pos)
{int res=0;while(pos>0){res+=num[pos];pos-=lowbit(pos);}return res;
}int main()
{while(scanf("%d",&n) && n){memset(num,0,sizeof num);int x,y;for(int i=1;i<=n;i++){scanf("%d%d",&x,&y);update(x,1);update(y+1,-1);}for(int i=1;i<=n;i++){printf("%d%s",getsum(i),i==n?"\n":" ");}}return 0;
}