大神题解:

http://blog.csdn.net/u014800748/article/details/47680899

The sum of gcd

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

You have an array ,the
length of  is

Let

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

First line has one integers

Second line has  integers

Third line has one integers ,the
number of questions

Next there are Q lines,each line has two integers ,

Output

For each question,you need to print

Sample Input

2
5
1 2 3 4 5
3
1 3
2 3
1 4
4
4 2 6 9
3
1 3
2 4
2 3

Sample Output

9
6
16
18
23
10

Author

SXYZ

Source

2015 Multi-University Training Contest 8

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int maxn=10100;
typedef long long int LL;

struct G
{
    G(){}
    G(int _id,LL _g):id(_id),g(_g){}

    int id;
	LL g;

	void toString()
	{
		printf("id: %d g: %lld\n",id,g);
	}
};

int n,a[maxn],Q;
vector<G> VL[maxn],VR[maxn];
struct Que
{
    int L,R,id;
    bool operator<(const Que& que) const
    {
        if(L!=que.L) return L<que.L;
        return R<que.R;
    }
}que[maxn];

void PreInit()
{
    /// get Left Point
    /// 以i为右端点,预处理出左边的段
    for(int i=1;i<=n;i++)
    {
        VL[i].clear();
        if(i==1)
        {
            VL[i].push_back(G(i,a[i]));
        }
        else
        {
			LL curg=a[i];int L=i;
			for(auto &it : VL[i-1])
			{
				int g=__gcd(it.g,curg);
				if(g!=curg) VL[i].push_back(G(L,curg));
				curg=g; L=it.id;
			}
			VL[i].push_back(G(L,curg));
        }
    }
    /// get Right Point
    /// 以i为左端点,预处理出右边的段
	for(int i=n;i>=1;i--)
	{
		VR[i].clear();
		if(i==n)
		{
			VR[i].push_back(G(i,a[i]));
		}
		else
		{
			LL curg=a[i];int R=i;
			for(auto &it : VR[i+1])
			{
				int g=__gcd(curg,it.g);
				if(g!=curg) VR[i].push_back(G(R,curg));
				curg=g; R=it.id;
			}
			VR[i].push_back(G(R,curg));
		}
	}
}

/// 计算L,R之间的值
LL calu(int type,int L,int R)
{
	LL ret=0;
	if(type==0)
	{
		int tr=R;
		for(auto &it : VL[R])
		{
			if(it.id>=L)
			{
				ret+=(tr-it.id+1)*it.g;
				tr=it.id-1;
			}
			else
			{
				ret+=(tr-L+1)*it.g;
				break;
			}
		}
	}
	else if(type==1)
	{
		int tr=L;
		for(auto &it : VR[L])
		{
			if(it.id<=R)
			{
				ret+=(it.id-tr+1)*it.g;
				tr=it.id+1;
			}
			else
			{
				ret+=(R-tr+1)*it.g;
				break;
			}
		}
	}
	return ret;
}

LL ans[maxn];

int main()
{
	int T_T;
	scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        PreInit();

		scanf("%d",&Q);
        for(int i=0,l,r;i<Q;i++)
        {
            scanf("%d%d",&l,&r);
            que[i].L=l; que[i].R=r; que[i].id=i;
        }
        sort(que,que+Q);

		int L=1,R=0; LL ret=0;
		for(int i=0;i<Q;i++)
		{
			while(R<que[i].R)
			{
				R++;
				ret+=calu(0,L,R);
			}
			while(R>que[i].R)
			{
				ret-=calu(0,L,R);
				R--;
			}
			while(L<que[i].L)
			{
				ret-=calu(1,L,R);
				L++;
			}
			while(L>que[i].L)
			{
				L--;
				ret+=calu(1,L,R);
			}
			ans[que[i].id]=ret;
		}

		for(int i=0;i<Q;i++)
			cout<<ans[i]<<endl;
    }
    return 0;
}