HDU 1556 Color the ball

Color the ball

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17276    Accepted Submission(s): 8640

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

 

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/*
每次更新a到b区间就可以了
*/
#include<iostream>
#include<string.h>
#include<stdio.h>
#define N 100010
using namespace std;
int c[N];
int n,a,b;
int lowbit(int x)
{
    return x&(-x);
}

void update(int x,int val)
{
    while(x<=n)
    {
        c[x]+=val;
        x+=lowbit(x);
    }
}

int getsum(int x)
{
    int s=0;
    while(x>0)
    {
        s+=c[x];
        x-=lowbit(x);
    }
    return s;
}

int main()
{
    //freopen("C:\Users\acer\Desktop\in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF&&n)
    {
        memset(c,0,sizeof c);
        //cout<<n<<endl;
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            //cout<<a<<" "<<b<<endl;
            update(a,1);//更新状态
            update(b+1,-1);//更新状态
        }
        //cout<<"ok"<<endl;
        for(int i=1;i<n;i++)
        {
            printf("%d ",getsum(i));
            //cout<<i<<endl;
        }    
        printf("%d
",getsum(n));
    }
    return 0;
}