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
", 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?'
':' ');
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?'
':' ');
}
return 0;
}