树状数组-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): 15579 Accepted Submission(s): 7766

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

题解:
依然是树状数组的初级应用,区间更新,单点求值。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <vector>
#include <string>
#include <set>
#include <utility>
const int MAXN=100005;
using namespace std;
long long int n,a[MAXN];
inline long long int LowBit(long long int x)
{
    return x&(-x);
}
inline void Add(long long int num,long long int x)
{
    while(num>0)
    {
        a[num]+=x;
        num-=LowBit(num);
    }
}
inline long long int GetVaule(long long int num)
{
    long long int sum=0;
    while(num<=n)
    {
        sum+=a[num];
        num+=LowBit(num);
    }
    return sum;
}
int main()
{
    long long int x,y;
    while(scanf("%lld",&n)!=EOF && n)
    {
        memset(a,0,sizeof(a));
        for(long long int i=1; i<=n; i++)
        {
            scanf("%lld %lld",&x,&y);
            Add(y,1);
            Add(x-1,-1);
        }
        for(long long int i=1; i<n; i++)
            printf("%lld ",GetVaule(i));
        printf("%lld\n",GetVaule(n));
    }
    return 0;
}

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