There are nn water tanks in a row, ii-th of them contains aiai liters of water. The tanks are numbered from 11 to nn from left to right.

You can perform the following operation: choose some subsegment [l,r][l,r] (1≤l≤r≤n1≤l≤r≤n), and redistribute water in tanks l,l+1,…,rl,l+1,…,r evenly. In other words, replace each of al,al+1,…,aral,al+1,…,ar by al+al+1+?+arr?l+1al+al+1+?+arr?l+1. For example, if for volumes [1,3,6,7][1,3,6,7] you choose l=2,r=3l=2,r=3, new volumes of water will be [1,4.5,4.5,7][1,4.5,4.5,7]. You can perform this operation any number of times.

What is the lexicographically smallest sequence of volumes of water that you can achieve?

As a reminder:

A sequence aa is lexicographically smaller than a sequence bb of the same length if and only if the following holds: in the first (leftmost) position where aa and bb differ, the sequence aa has a smaller element than the corresponding element in bb.

Input
The first line contains an integer nn (1≤n≤1061≤n≤106) — the number of water tanks.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106) — initial volumes of water in the water tanks, in liters.

Because of large input, reading input as doubles is not recommended.

Output
Print the lexicographically smallest sequence you can get. In the ii-th line print the final volume of water in the ii-th tank.

Your answer is considered correct if the absolute or relative error of each aiai does not exceed 10?910?9.

Formally, let your answer be a1,a2,…,ana1,a2,…,an, and the jury’s answer be b1,b2,…,bnb1,b2,…,bn. Your answer is accepted if and only if |ai?bi|max(1,|bi|)≤10?9|ai?bi|max(1,|bi|)≤10?9 for each ii.

Examples
Input
4
7 5 5 7
Output
5.666666667
5.666666667
5.666666667
7.000000000
Input
5
7 8 8 10 12
Output
7.000000000
8.000000000
8.000000000
10.000000000
12.000000000
Input
10
3 9 5 5 1 7 5 3 8 7
Output
3.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
5.000000000
7.500000000
7.500000000
Note
In the first sample, you can get the sequence by applying the operation for subsegment [1,3][1,3].

In the second sample, you can’t get any lexicographically smaller sequence.
思路：一開始我們將每一個(gè)數(shù)字當(dāng)做一個(gè)區(qū)域，從后往前遍歷，如果當(dāng)前區(qū)域加上之前的區(qū)域可以使得平均值變小的話，那么我們就合并這兩個(gè)區(qū)域，這樣出來就是最優(yōu)的了。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=1e6+100; struct node{ll num;ll cnt; }p[maxx]; ll a[maxx]; int n;int main() {scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%lld",&a[i]);int fro=0;p[++fro].num=a[n];p[fro].cnt=1ll;for(int i=n-1;i>=1;i--){ll cnt=1ll;ll sum=a[i];while(fro&&p[fro].num*(p[fro].cnt+cnt)<p[fro].cnt*(p[fro].num+sum)){cnt+=p[fro].cnt;sum+=p[fro].num;--fro;}p[++fro].cnt=cnt;p[fro].num=sum;}for(int i=fro;i>=1;i--){for(int j=1;j<=p[i].cnt;j++) printf("%.9lf\n",(double)p[i].num/(double)p[i].cnt);}return 0; }

努力加油a啊，(^o)/~

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