Bzoj4488 [Jsoi2015] Greatest common divisor

Time Limit:10 Sec Memory limit:256 MB
submit:172 solved:101

Description

Given a sequence of positive integers of length N ai for any successive subsequence of it
{al,al+1...ar}, we define its weight w (l,r) as the product of its length with the greatest common divisor of all elements in the sequence, i.e. w (l,r) = (r-l+1) ∗gcd (Al: AR).
Jyy want to find out which sub-sequence is the most weighted.

Input

The input line consists of a positive integer N.
The next line, containing n positive integers, represents the sequence AI
1 < = Ai < = 10^12, 1 < = N < = 100,000

Output

The output file contains a single positive integer that represents the weight of the subsequence with the highest weight.

Sample Input5
theSample Output80
The best sub-sequence is a subsequence of the last 4 elements. Hintsource

Scanning water Problems

It's time for the water problem again ~

The sequence of gcd converges quickly, as if it were concluded that different gcd would not exceed the log (sequence length) of

Use a map to save the current GCD where it first appeared, scan the sequence of statistical answers.

If that conclusion is correct, the complexity is probably $ O (n log^2 N) $ (map comes with a log)

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cmath>5#include <cstring>6#include <map>7 #defineLL Long Long8 using namespacestd;9 Const intmxn=100010;Ten LL Read () { OneLL x=0, f=1;CharCh=GetChar (); A      while(ch<'0'|| Ch>'9'){if(ch=='-') f=-1; ch=GetChar ();} -      while(ch>='0'&& ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} -     returnx*F; the } -ll GCD (ll A,ll b) {returnB?GCD (b,a%b): A;} -Map<ll,int>mp,tmp; -Map<ll,int>:: iterator it; + intN; - LL A[MXN]; + intMain () { A //freopen ("In.txt", "R", stdin); at     inti,j; -n=read (); -LL ans=0; -      for(i=1; i<=n;i++){ -A[i]=read (); ans=Max (ans,a[i]); -          for(It=mp.begin (); It!=mp.end (); it++){ inLL G=GCD (*it). First,a[i]); -Ans=max (ans,g* (i-(*it). second+1)); to             if(!tmp.count (g)) tmp[g]= (*it). Second; +             ElseTmp[g]=min (Tmp[g], (*it). second); -         } the         if(!tmp.count (A[i])) tmp[a[i]]=i; *mp=tmp; $ tmp.clear ();Panax Notoginseng     } -printf"%lld\n", ans); the     return 0; +}

Bzoj4488 [Jsoi2015] Greatest common divisor