二分 + 差分

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 V_i 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 T_i, each pile will reduce its volume by T_i. 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, ..., V_N (0 ≤ V_i ≤ 10⁹), where V_i is the initial size of a snow pile made on the day i.

The third line contains N integers T₁, T₂, ..., T_N (0 ≤ T_i ≤ 10⁹), where T_i 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.

Example

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 ： 输入的一个 n ，表示天数，然后输入的第一行表示在这天产生的雪的量，然后再一行表示雪在这天的融化量是多少，对于当前的一天，他不仅会融化当天的，还会融化前面所有天的雪，最后输出每天融化雪的量。

思路分析 ：最开始的想法是纯用线段树去搞它，写了一会交上去WA了... 后来问了下大佬的思路，才知道原来这题是去二分雪的量，去判断雪在哪天融化 。然后差分一遍就没问题了。

代码示例 ：

#define ll long long
const ll maxn = 1e5+5;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;

ll v[maxn], te[maxn];
ll sum[maxn];
ll ans[maxn], pp[maxn], f[maxn];
ll start;

bool check(ll p, ll vi){
    ll ss = sum[p] - sum[start-1];
    if (ss >= vi) return true;
    else return false;
}

void fun(ll l, ll r, ll vi){
    ll mid, pos;
    while(l <= r){
        mid = (l+r) >> 1;
        if (check(mid, vi)) {pos = mid; r = mid-1;} 
        else l = mid+1;
    }
    //printf("mid = %lld\n", pos);
    ans[pos]--;
    ans[start]++;
    pp[pos] += vi-(sum[pos-1]-sum[start-1]);
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll n;
    cin >> n;
    for(ll i = 1; i <= n; i++) scanf("%d", &v[i]);
    for(ll i = 1; i <= n; i++) {
        scanf("%lld", &te[i]);
        sum[i] = sum[i-1] + te[i];
    }
    sum[n+1] = sum[n] + 10000000000;
    for(ll i = 1; i <= n; i++){
        start = i;
        fun(i, n+1, v[i]); 
    }
    for(ll i = 1; i <= n; i++) {
        
        ans[i] += ans[i-1];
        f[i] = ans[i]*te[i] + pp[i];
    }
    for(ll i = 1; i <= n; i++) printf("%lld%c", f[i], i==n?'\n':' ');
    return 0;
}

 

方法二 ： 优先队列 + 前缀和

看了大佬们的代码，发现优先队列也可以，每次将当天前面所有的温度和累加再加上当前天的雪的总量，这样打入优先队列的雪量就可以视为是从同一天打入的

代码示例 ：

#define ll long long
const ll maxn = 1e5+5;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;

ll v[maxn], te[maxn];
ll sum[maxn];
priority_queue<ll, vector<ll>, greater<ll> >que;

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    ll n;
    cin >> n;
    for(ll i = 1; i <= n; i++) scanf("%d", &v[i]);
    for(ll i = 1; i <= n; i++) {
        scanf("%lld", &te[i]);
        sum[i] = sum[i-1] + te[i];
    }
    ll pt = 0;
    for(ll i = 1; i <= n; i++){
        que.push(sum[i-1]+v[i]);
        pt += te[i];
        ll ans = 0;
        
        while(!que.empty()){
            ll x = que.top();
            if (x > pt) break;
            que.pop();
            ans += x-(pt-te[i]);    
        }
        ans += que.size()*te[i];
        printf("%lld%c", ans, i==n?'\n':' ');
    }    
    return 0;
}