HDU5753 permutation Bo (2016 multi-school training)

permutation Bo

Time limit:2000/1000 MS (java/others) Memory limit:131072/131072 K (java/others)
Total submission (s): 777 Accepted Submission (s): 468
Special Judge

problem DescriptionThere is sequencesh1∼hn andC1∼cn .h1∼hn is a permutation of1∼n . Particularly,h0=hn+1=0 .

We define the expression[condition] is 1 whencondition is True,is 0 whencondition Is False.

Define the functionF(H)=∑NI=1CI[HI>HI−1  a nd   hi>h i +1]

Bo has gotten the value ofc1∼< Span id= "mathjax-span-142" class= "Msubsup" >cn , and he wants to know the expected value of f (h) . InputThis problem have multi test cases (no more than A).

For each test case, the first line contains a non-negative integern(1≤n≤) , second line containsn non-negative Integer ci ( Span id= "mathjax-span-173" class= "mn" >0≤ci≤1000 ) /span> .  OutputFor each test cases print a decimal-the expectation ofF(h).

If the absolute error between your answer and the standard answer are no more than −4, your solution'll be accepted.  Sample Input43 2 4 553 5 Sample Output6.00000052.833333 title Link: http://acm.hdu.edu.cn/showproblem.php?pid=5753 depending on the test instructions, you can consider each location's contribution to your expectations. When I is not at the end of the arrangement, there are 3! = 6 size relationships, of which 2 have contributed to the expectation, so the contribution is CI/3; When I is at both ends, there are 2! = 2 size relationships, of which 1 have contributed to the expectation, so the contribution is CI/2. (Note the case of n = = 1)

1#include <iostream>2#include <cstdio>3#include <algorithm>4#include <cstring>5#include <string>6 using namespacestd;7typedefLong Longll;8typedef unsignedLong Longull;9typedefLong Doubleld;Ten #defineMAX 1005 One  A intN; - intC[max]; - Doubleans; the voidSolve () - { -      for(intI=1; i<=n;i++) -CIN >>C[i]; +     if(n==1) -     { +Ans = c[1]; A         return; at     } -     if(n==2) -     { -Ans = (c[1]+c[2])/2.0; -         return; -     } inAns =0; -      for(intI=2; i<n;i++) toAns + = (c[i])/3.0; +ans+=c[1]/2.0; -ans+=c[n]/2.0; the } * intMain () $ {Panax Notoginseng      while(~SCANF ("%d",&N)) -     { the solve (); +printf"%.6f\n", ans); A     } the     return 0; +}

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HDU5753 permutation Bo (2016 multi-school training)