HDU 4066 2011ACM福州网络赛

最新推荐文章于 2025-11-26 15:59:01 发布

最新推荐文章于 2025-11-26 15:59:01 发布 · 179 阅读

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#java

本文深入探讨了程序设计和算法优化的关键技巧，包括数据结构、算法策略、版本控制、项目管理工具等核心主题，旨在提升开发效率和代码质量。

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; }