Codeforces Round #470 div2 C
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.
題意 每天會有 ai 個雪球被建立 每天會有bi 個雪球會被融化掉 求每天的融化的 個數
md 好難啊 線段樹+二分+字首和
#include<bits/stdc++.h>
using namespace std;
#define maxn 1000000+1000
#define LL long long
LL tre[maxn*10]={0};
LL a[maxn],b[maxn],c[maxn];
void up(LL in,LL l,LL r,LL x,LL y){
if(x>y)return;
if(l==x&&r==y){tre[in]++;return;}
long long mid=(l+r)/2;
if(y<=mid)up(in*2,l,mid,x,y);
else if(x>mid)up(in*2+1,mid+1,r,x,y);
else up(in*2,l,mid,x,mid),up(in*2+1,mid+1,r,mid+1,y);
}
LL que(LL in,LL l,LL r,LL x){
if(l==r)return tre[in];
long long mid=(l+r)/2;
if(mid>=x)return que(in*2,l,mid,x)+tre[in];
return que(in*2+1,mid+1,r,x)+tre[in];
}
int main(){
LL n;
cin>>n;
for(LL j=1;j<=n;j++){
scanf("%lld",&a[j]);
}
b[0]=0;c[0]=0;
for(LL k=1;k<=n;k++){
LL x;
scanf("%lld",&x);
b[k]=b[k-1]+x;
}
for(LL j=1;j<=n;j++){
a[j]+=b[j-1];
LL i=1LL*(lower_bound(b+j,b+n+1,a[j])-b);
up(1,1,n,j,i-1);
c[i]+=a[j]-b[i-1];
}
for(LL j=1;j<=n;j++){
c[j]+=que(1,1,n,j)*(b[j]-b[j-1]);
cout<<c[j]<<" ";
}
return 0;
}
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