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.
题目大意
有个人每天都会弄一堆雪在院子里面，然后天气很热，每天雪都会融化当前温度的体积，直到这堆雪融化完了，问你每天会融化多少雪。
思路
这个题目开始想多了，很烦，其实还是比较简单的，就求出每堆雪会在哪天融化完的，还有融化当天比平常少融化多少，减去就好了。但是在求哪天融化的时候，必须优化一点，否则时间不够，优化的时候，后面加入的雪堆就当做一开就存在，这样就只要一次扫描就可以求出所有的雪堆哪天融化的，用了一下lower_bound()这个函数，可以快速求出雪堆是哪天融化的.
代码
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <string>
#include <vector>
#include <stack>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#define ll long long
using namespace std;
const int maxn = 100005;
struct node
{
ll n;
ll total;
} pi[maxn];
ll v[maxn], t[maxn];
ll num[maxn];
void init()
{
for (int i = 0; i < maxn; i++)
{
pi[i].n = 0;
pi[i].total = 0;
v[i] = 0;
t[i] = 0;
num[i] = 0;
}
}
int main()
{
freopen("test.txt", "r", stdin);
// freopen("test.out","w",stdout);
int n;
while (~scanf("%d", &n))
{
init();
for (int i = 1; i <= n; i++)
{
scanf("%lld", &v[i]);
}
for (int i = 1; i <= n; i++)
{
scanf("%lld", &t[i]);
v[i] += num[i - 1];
num[i] = num[i - 1] + t[i];
}
for (int i = 1; i <= n; i++)
{
if (v[i] > 0)
{
int a = lower_bound(num, num + n + 1, v[i]) - num;
if (num[a] >= v[i])
{
if (pi[a].n > 0)
{
pi[a].n++;
pi[a].total += num[a] - v[i];
}
else
{
pi[a].n = 1;
pi[a].total = num[a] - v[i];
}
}
}
else
{
pi[i].n++;
pi[i].total += t[i];
}
}
int cou = 0;
for (int i = 1; i <= n; i++)
{
printf("%lld ", t[i] * (i - cou) - pi[i].total);
cou += pi[i].n;
}
printf("\n");
}
return 0;
}