Color the ball
Time Limit: 9000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29305 Accepted Submission(s): 14264
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
Author
8600
Source
//hdu 1556 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <set> #include <map> using namespace std; #define ll long long #define lowbit(x) x&(-x) const int N=1e5+9; ll a[N],c[N]; ll n,x,y; void update(ll x,ll num){ while(x<=n){ c[x]+=num; x+=lowbit(x); } } ll getsum(ll x){ ll sum = 0; while(x>0){ sum+=c[x]; x-=lowbit(x); } return sum; } int main() { while(~scanf("%lld",&n),n){ for(ll i = 1 ;i <=n+9;i++){ a[i]=0; c[i] =0; } for(ll i =0;i<n;i++){ scanf("%lld%lld",&x,&y); update(x,1); update(y+1,-1); } for(ll i =1;i<=n;i++){ printf("%lld%c",getsum(i),i==n?‘\n‘:‘ ‘); } } return 0; }
原文地址:https://www.cnblogs.com/tingtin/p/9902412.html
时间: 2024-10-13 11:31:37