Codeforces 470Div 2 C. Producing Snow

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本文介绍了一个基于线段树的数据结构算法问题，通过模拟每天雪堆的融化情况来计算总融化体积。输入包括天数、初始雪堆大小及每日温度，输出为每天的雪融化总量。

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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.

算出第i堆学在第j天全部消融，算出第i堆雪在j天的消融量。再用线段树维护从第i天道第j-1天的消融量

#include<bits/stdc++.h>
#define ll long long
#define maxn 100010
using namespace std;
ll A[maxn<<2];
ll a[maxn],b[maxn],c[maxn];
void update(int k,int L,int R,int l,int r)
{
    if(r<l)
        return;
    if(l==L&&r==R)
    {
        A[k]++;
        return;
    }
    ll mid=(L+R)/2;
    if(r<=mid)
        update(2*k,L,mid,l,r);
    else if(l>mid)
        update(2*k+1,mid+1,R,l,r);
    else
    {
        update(2*k,L,mid,l,mid);
        update(2*k+1,mid+1,R,mid+1,r);
    }
}
ll query(int k,int L,int R,int t)
{
    if(L==R)
        return A[k];
    ll mid=(L+R)/2;
    if(t<=mid)
        return query(2*k,L,mid,t)+A[k];
    else
        return query(2*k+1,mid+1,R,t)+A[k];
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++)
    {
        int t;
        scanf("%d",&t);
        b[i]=b[i-1]+t;
    }
    for(int i=1;i<=n;i++)
    {
        a[i]+=b[i-1];//先把i天之前的温度加上
        ll j=1ll*(lower_bound(b+i,b+n+1,a[i])-b);//找到第i堆雪存在的天数
        c[j]+=a[i]-b[j-1];//计算出第j天第i堆雪的消融量
        update(1,1,n,i,j-1);//线段树维护第i天到第j-1天雪量大于等于温度的雪堆数目
    }
    for(int i=1;i<=n;i++)
    {
        c[i]+=query(1,1,n,i)*(b[i]-b[i-1]);
        printf("%lld ",c[i]);
    }
    return 0;
}