题目：
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
Input
3 10 10 5 5 7 2
Output
5 12 4
Input
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.
题意：给我们 n 天的时间，每天都会有新的雪堆生成，然后每天又会有雪堆消耗，当然每天的消耗是不同的，叫我们输出每天的雪量的消耗。
解题思路：上来确实是很简单，直接暴力求解试一下，但是 很光荣的 T 了，之后一直卡，不知道进展，后来训练结束后，大神的讲解我也没有听的很明白，后来思考了好久，今天终于有些领悟了，来补一下这道题目。
咱们可以对数据进行一下操作，使每天都是第一天，然后就对他们进行操作，看看累计消耗的值，能够撑多久，（优先队列的长度）如果满足条件 ，那么就ans加和 ，（即前边几天都符合的时候），不符合的时候，就说明前边生成的雪堆不足以支撑这些天的消耗了，咱们就从头开始减 直到雪堆完全消耗。
具体啥的见代码
ac代码：
#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
priority_queue<long long int ,vector<long long int >,greater<long long int > >que;
const int N=1e5+5;
int num[N];
long long int res[N];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
long long int ans;
res[0]=0;
for(int i=1;i<=n;i++)
{
ans=0;
scanf("%I64d",&res[i]);
res[i]+=res[i-1];//累计前几天 的消耗和
que.push(num[i]+res[i-1]);//将今天产生的和昨天消耗的加在一起，更改为第一天的状态
while(!que.empty()&&res[i]>=que.top())
{//如果队列内的雪堆撑不住消耗的话，就只加上不满足条件的雪堆的剩余
ans+=que.top()-res[i-1];
que.pop();
}
ans+=que.size()*(res[i]-res[i-1]);//撑过了几天
printf("%I64d ",ans);
}
printf("\n");
}
return 0;
}