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。
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
#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;
}Keep it simple!

浙公网安备 33010602011771号