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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
31 12 23 331 11 21 30
 

 

Sample Output
1 1 13 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 100010using 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\n",getsum(n));    }    return 0;}

 

HDU 1556 Color the ball