hdu 4923 Room and Moor

Room and Moor

Time Limit: 12000/6000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

Total Submission(s): 539    Accepted Submission(s): 147


Problem Description

PM Room defines a sequence A = {A1, A2,..., AN}, each of which is either 0 or 1. In order to beat him, programmer Moor has to construct another sequence B = {B1, B2,... , BN} 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 Ai.

 

Output

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

 

Sample Input

4
9
1 1 1 1 1 0 0 1 1
9
1 1 0 0 1 1 1 1 1
4
0 0 1 1
4
0 1 1 1

 

Sample Output

1.428571
1.000000
0.000000
0.000000

 

Author

BUPT

 

官方题解：



代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef __int64 ll;
struct node
{
double l,r;
double p;
} b[50010];

int main()
{
int T;
cin>>T;
int n;
int s[100100];
while(T--)
{
cin>>n;
double ans=0;
bool flag=true;
for(int i=0; i<n; i++)
{
cin>>s[i];
if(s[i]==0&&flag) i--,n--;
else flag=false;
}
int l=0,r=n-1;
for(; r>=0; r--)
if(s[r]!=1)
{
break;
}
int k=1;
b[0].p=0;
b[1].l=0,b[1].r=0;
for(int i=l; i<=r; i++)
{
if(s[i]==1) b[k].l++;
else b[k].r++;

b[k].p=b[k].l/(b[k].l+b[k].r);
while(b[k].p<b[k-1].p)
{
b[k-1].l+=b[k].l;
b[k-1].r+=b[k].r;
b[k-1].p=b[k-1].l/(b[k-1].l+b[k-1].r);
b[k].l=0,b[k].r=0;
k--;
}
k++;
b[k].l=0,b[k].r=0;
}
for(int i=1; i<k; i++)
{
ans+=b[i].r*b[i].p;
}
printf("%.6lf\n",ans);
}
return 0;
}