Producing Snow

C.

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Alice likes snow a lot! Unfortunately, this year’s winter is already over, and she can’t expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.

Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other.

Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.

You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day.

Input
The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days.

The second line contains N integers V1, V2, …, VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.

The third line contains N integers T1, T2, …, TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.

Output
Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.

Examples
inputCopy
3
10 10 5
5 7 2
output
5 12 4
inputCopy
5
30 25 20 15 10
9 10 12 4 13
output
9 20 35 11 25
Note
In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
ac代码如下：

//思路：所有堆如果均放在第一天的话，这个堆将是多大v[i]+sum;sum+=t[i];
//把所有堆放在一个多重集合或者优先队列中，清楚雪如果*m.begin()<=sum;
//每个堆都被添加或者移除一次，因此时间复杂度为N*logN；
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<ctype.h>
#include<stack>
#include<math.h>
#include <string>
#include<algorithm>
#include <iterator>
#include <stdlib.h>

using namespace std;

typedef unsigned long long ULL;

int main()
{
    int v[100005],t[100005];
    int n;
    ULL sum;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&v[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&t[i]);
    multiset<ULL> m;
    sum=0;
    ULL now=0;
    for(int i=1;i<=n;i++)
    {
        now=0;
        m.insert(v[i]+sum);
        sum+=t[i];
        while(!m.empty()&&*m.begin()<=sum)
        {
            now+=*m.begin()-(sum-t[i]);
//            cout<<"now="<<now<<endl;
            m.erase(m.begin());
        }
        now+=t[i]*m.size();
        printf("%I64d ",now);
    }
    printf("\n");
    return 0;
}