UVA 1642 Magical GCD 暴力+簡單數論

枚舉右端點,往前查找左端點....

右端點一定的話,最多只有log個不同的gcd值,

用一個數組記錄不同的GCD的值,對每個相同的GCD值記錄一下最靠左的位置...

因爲GCD值不是很多所以 移動右端點時暴力統計即可..

對與樣例:
30 60 20 20 20

從第1個數座右端點開始枚舉  // (gcd,位置)

(30,1)

枚舉以第2個數做爲右端點

(30,1) (60,2)

第3個數

(10,1)  (20,2)

....

後面幾個數都是一樣的...

第5個數

(10,1)  (20,2)

這時 20*(5-2+1) = 80 最大

点击打开链接

1642 Magical GCDThe Magical GCD of a nonempty sequence of positive integers is defined as the product of its lengthand the greatest common divisor of all its elements.Given a sequence (a1, . . . , an), find the largest possible Magical GCD of its connected subsequence.InputThe first line of input contains the number of test cases T. The descriptions of the test cases follow:The description of each test case starts with a line containing a single integer n, 1 ≤ n ≤ 100000.The next line contains the sequence a1, a2, . . . , an, 1 ≤ ai ≤ 1012.OutputFor each test case output one line containing a single integer: the largest Magical GCD of a connectedsubsequence of the input sequence.Sample Input1530 60 20 20 20Sample Output80

/* ***********************************************
Author        :CKboss
Created Time  :2015年02月04日 星期三 13时12分27秒
File Name     :UVA1642.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;
const int maxn = 101000;

struct YUE
{
	LL val,pos;
};

bool cmp(YUE a,YUE b)
{
	if(a.val!=b.val) return a.val<b.val;
	return a.pos<b.pos;
}

LL n;
LL a[maxn];

LL gcd(LL a,LL b)
{
	return (b)?gcd(b,a%b):a;
}

vector<YUE> yue,temp;

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    
	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		yue.clear(); temp.clear();
		LL ans=0;
		scanf("%lld",&n);
		for(int i=0;i<n;i++)
		{
			scanf("%lld",a+i);
			ans=max(ans,a[i]);
		}

		yue.push_back((YUE){a[0],0});

		for(int i=1;i<n;i++)
		{
			int sz=yue.size();
			temp.clear();
			for(int j=0;j<sz;j++)
			{
				LL g=gcd(a[i],yue[j].val);
				temp.push_back((YUE){g,yue[j].pos});
			}
			temp.push_back((YUE){a[i],i});
			sort(temp.begin(),temp.end(),cmp);
			yue.clear(); LL last=-1;
			for(int j=0,si=temp.size();j<si;j++)
			{
				if(j==0) 
				{
					last = temp[j].val;
					yue.push_back(temp[j]);
				}
				else
				{
					if(last==temp[j].val) continue;
					else
					{
						last = temp[j].val;
						yue.push_back(temp[j]);
					}
				}
			}
			sz=yue.size();
			for(int j=0;j<sz;j++)
			{
				ans=max(ans,yue[j].val*(i-yue[j].pos+1LL));
			}
		}
		printf("%lld\n",ans);
	}

    return 0;
}