hdu 4923 第六次多校

Room and Moor

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

Total Submission(s): 79 Accepted Submission(s): 12



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

Source

2014 Multi-University Training Contest 6



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贪心即可 用栈维护区间递增，如果比栈顶小，就合并区间，直到合法为止；

然后每个区间f(a,b)=x*y/(x+y), x代表区间内1的个数，y代表区间内0的个数。

代码：

#include <cstdio>
#include <stack>
using namespace std;
struct Node {
    int ae, be;
    double ce;
    void init()
    {
        ce=double(be)/double(ae+be);
    }
};
int main ()
{
    //freopen("data.in","r",stdin);
    int n,T,date,oldDate;
    int BB;
    int BH;
    int a,b;
	double ans;
	Node temp;
	stack<Node> q;
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d",&n);
        a=0,b=0,BB=0,BH=0;
        oldDate=1;
        for (int i=0;i<n;i++)
        {
            scanf("%d",&date);
            if((date!=oldDate)&&i!=0)
            {
                temp.ae=a;
                temp.be=b;
                temp.init();
                while (!q.empty()&&temp.ce<q.top().ce)
                {
                    temp.ae=temp.ae+q.top().ae;
                    temp.be=temp.be+q.top().be;
                    temp.init();
                    q.pop();
                }
                q.push(temp);
                if(date==0)
                {
                    a=1;
                    b=0;
                }
                else {a=0;
                    b=1;}
            }
            else {
                if(date==0)
                {
                    a++;
                }
                else {
                    b++;
                }
            }
            if((i==n-1)&&date==0)
            {
                temp.ae=a;
                temp.be=b;
                temp.init();
                while (!q.empty()&&temp.ce<q.top().ce)
                {

                    temp.ae=temp.ae+q.top().ae;
                    temp.be=temp.be+q.top().be;
                    temp.init();
                    q.pop();
                }
                q.push(temp);
            }
        
            oldDate=date;
        
        }
        ans=0;
        while (!q.empty())
        {
            ans+=q.top().ce*q.top().ae;
            q.pop();
        }
        printf("%.6f\n",ans);
    }
    return 0;
    
    
}