HDU 4066 2011ACM福州网络赛

Random Sequence

Time Limit: 3000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 98Accepted Submission(s): 51

Problem Description

There is a random sequence L whose element are all random numbers either -1 or 1 with the same possibility. Now we define MAVS, the abbreviate of Maximum Absolute Value Subsequence, to be any (if more than one) subsequences of L whose absolute value is maximum among all subsequences. Given the length of L, your task is to find the expectation of the absolute value of MAVS.

Input

There is only one input file. The first line is the number of test cases T. T positive integers follow, each of which contains one positive number not greater than 1500 denoted the length of L.

Output

For each test case, output the expectation you are required to calculate. Answers are rounded to 6 numbers after the decimal point.(as shown in the sample output)

Sample Input

3
1
5
10

Sample Output

Case 1: 1.000000
Case 2: 2.750000
Case 3: 4.167969

Source

The 36th ACM/ICPC Asia Regional Fuzhou Site —— Online Contest

我也不知道怎么做

反正是写了个暴力的程序，找规律过掉的。。嘿嘿

代码：

#include<stdio.h>

double ans[1505];

int main()
{
	int i,n;
	double x=1.0;
	ans[1]=1.0;
	for(i=2;i<=1500;i++)
	{
		if(i%2==0)
			x=x*(i-1.0)/(i*1.0);
		ans[i]=ans[i-1]+x;
	}
	int t,T;
	scanf("%d",&T);
	for(t=1;t<=T;t++)
	{
		scanf("%d",&n);
		printf("Case %d: %.6lf\n",t,ans[n]);
	}
	return 0;
}

暴力的打表程序：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
main(){
	 int i,j,k,m=1,temp,temp1,temp2,Max1,Max2,len;
	 double a[100];
 	 for(i=1;i<=26;i++){
 	     len=0;
	     m*=2;
	     for(j=0;j<m;j++){
     		 temp1=temp2=Max1=Max2=0;
     		 temp=j;
  	         for(k=1;k<=i;k++){
 		     	 if(temp%2==0) {temp1++;temp2--;}
 		         else {temp1--;temp2++;}
	             if(temp1<0) temp1=0;
	             if(temp2<0) temp2=0;
	             temp/=2;
				 if(Max1<temp1) Max1=temp1;
	             if(Max2<temp2) Max2=temp2;
			 }
			 len+=Max1>Max2?Max1:Max2;
     	 }
         printf("%d: %d %d %.6lf\n",i,len,m,(len*1.0)/(m*1.0));
		 a[i]=(len*1.0)/(m*1.0);
	 }
	 for(i=2;i<=26;i++)
		 printf("%.15lf\n",a[i]-a[i-1]);
	 return 0;
}