HDOJ 5381 The sum of gcd 莫队算法

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)
Total Submission(s): 526    Accepted Submission(s): 226

Problem Description

You have an array A,the length of A is n
Let f(l,r)=∑ri=l∑rj=igcd(ai,ai+1....aj)

 

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 n
Second line has n integers Ai
Third line has one integers Q,the number of questions
Next there are Q lines,each line has two integers l,r
1≤T≤3
1≤n,Q≤104
1≤ai≤109
1≤l<r≤n

 

Output

For each question,you need to print f(l,r)

 

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
",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;
}