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.
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 a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.
3 10 10 5 5 7 2
5 12 4
5 30 25 20 15 10 9 10 12 4 13
9 20 35 11 25
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.
这里我是暴力做的，稍微优化一下，900ms+过了
存储所有温度的总合，遍历的时候判断当前雪是否大于剩余所有温度和，如果是的话，就舍去，计数器加一。
用优先队列也可以。我这里用的是vector
下面是ac代码
#include<bits/stdc++.h>
using namespace std;
#define MID(x,y) ((x+y)>>1)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);
const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int N=2e5+7;
int a[N],b[N], c[N];
vector<long> v;
int main()
{
int n, i ,j ,k, f = 0;
__int64 sumT = 0;
scanf("%d", &n);
for (i = 0 ; i < n ; i ++) {
scanf("%d", &a[i]);
}
for (i = 0 ; i < n ; i ++) {
scanf("%d", &b[i]);
sumT += b[i];
}
int cnt = 0;
__int64 sum;
for (i = 0; i < n ; i ++) {
sum = 0;
if (a[i] <= b[i]) {
sum += a[i];
} else {
if (a[i] >= sumT) {
++cnt;
} else {
v.push_back(a[i]);
}
}
sumT -= b[i];
int size = v.size();
for (int j = size - 1; j >= 0 ; j --) {
if (v[j] <= b[i]) {
sum += v[j];
v.erase(v.begin() + j);
} else {
sum += b[i];
v[j] -= b[i];
}
}
sum += (__int64)b[i] * cnt;
if (i == 0)
printf("%I64d", sum);
else printf(" %I64d", sum);
}
printf("\n");
}