HDOJ 4923 Room and Moor

用一个栈维护b的值，每次把一个数放到栈顶。看栈首的值是不是大于这个数，如果大于的话将栈顶2个元素合并，b的值就是这两个栈顶元素的平均值。。。

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 943    Accepted Submission(s): 291

Problem Description

PM Room defines a sequence A = {A₁, A₂,..., A_N}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B₁, B₂,... , B_N} of the same length, which satisfies that:

 

Input

The input consists of multiple test cases. The number of test cases T(T<=100) occurs in the first line of input.

For each test case:
The first line contains a single integer N (1<=N<=100000), which denotes the length of A and B.
The second line consists of N integers, where the ith denotes A_i.

 

Output

Output the minimal f (A, B) when B is optimal and round it to 6 decimals.

 

Sample Input

491 1 1 1 1 0 0 1 191 1 0 0 1 1 1 1 140 0 1 140 1 1 1

 

Sample Output

1.4285711.0000000.0000000.000000

 

Author

BUPT

 

Source

2014 Multi-University Training Contest 6

 

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stack>#include <cmath>using namespace std;const double eps=1e-8;double a[110000],b[110000];typedef pair<double,double> pDD;int n;stack<pDD> STACK;int main(){int T_T;scanf("%d",&T_T);while(T_T--){scanf("%d",&n);while(STACK.size()) STACK.pop();for(int i=0;i<n;i++){scanf("%lf",a+i);pDD A=pDD(a[i],1);while(STACK.size()&&A.first+eps<STACK.top().first){pDD B=STACK.top(); STACK.pop();double Sec=A.second+B.second;double Fst=(A.first*A.second+B.first*B.second)/Sec;A.first=Fst; A.second=Sec;}STACK.push(A);}int now=n-1;while(STACK.size()){pDD u=STACK.top(); STACK.pop();int sz=u.second;for(int i=now,j=0;i>=0&&j<sz;i--,j++){b[now--]=u.first;}}double ans=0;for(int i=0;i<n;i++){ans+=(a[i]-b[i])*(a[i]-b[i]);}printf("%.6lf\n",ans);}return 0;}