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; }