hdoj_1556Color the ball

Color the ball

Time Limit: 9000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5330    Accepted Submission(s): 2841


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
改段求点。


#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
#pragma warning(disable : 4996)

const int MAXN = 100005;
int tree[MAXN];
int n;

int LowBit(int t)
{
	return t&(-t);  
}

void Update(int pos, int num)
{
	while(pos <= n)
	{
		tree[pos] += num;
		pos += LowBit(pos);
	}
}

int GetSum(int end)
{
	int sum = 0;
	while(end > 0)
	{
		sum += tree[end];
		end -= LowBit(end);
	}
	return sum;
}

int main()
{
	freopen("in.txt","r",stdin);
	int x, y;
	while(scanf("%d", &n) != EOF)
	{
		if(n == 0)
		{
			break;
		}
		memset(tree, 0, sizeof(tree));
		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 ", GetSum(i));
		}
		printf("%d\n", GetSum(n));
	}
	return 0;
}


原文地址:https://www.cnblogs.com/lgh1992314/p/5834986.html