HDU 5381(The sum of gcd-莫队算法解决区间段gcd的和)

The sum of gcd

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 784    Accepted Submission(s): 335

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

 

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预处理一段的gcd

我们发现一段的gcd是一样的

1 2 3 6 6 6 12

1 1 3 6 6 6 12 //从12开始向左的gcd

显然最多有logN段，接下来用莫队算法+杨氏转移 

O(nlogn+nsqrt(n)logn)=O(n^1.5*logn)

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (10000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,a[MAXN],Q;
struct seg {
	int l,r,i;
	friend bool operator<(seg a,seg b){ return (int)((a.l)/sqrt(n))^(int)((b.l)/sqrt(n))?(int)((a.l)/sqrt(n))<(int)((b.l)/sqrt(n)):a.r<b.r;
	}
}comm[MAXN];
ll ans[MAXN];

ll gcd(ll a,ll b){if (b==0) return a; return gcd(b,a%b);}

int h[MAXN][100]={0};
ll val[MAXN][100]={0};
void init(int h[][100],ll  val[][100])
{
	MEM(val) MEM(h)
	For(i,n) h[i][0]=0;
	h[1][0]=1; h[1][1]=1; val[1][1]=a[1];
	Fork(i,2,n) {
		int &k=h[i][0];
		val[i][++k]=a[i];h[i][k]=i;
		For(j,h[i-1][0]) {
			ll p=gcd(val[i][k],val[i-1][j]);
			if (p!=val[i][k]) 
				val[i][++k]=p;
			h[i][k]=h[i-1][j];
		}
	}
}

int h2[MAXN][100]={0};
ll val2[MAXN][100]={0};
void init2(int h[][100],ll  val[][100])
{
	MEM(val) MEM(h)
	For(i,n) h[i][0]=0;
	h[n][0]=1; h[n][1]=n; val[n][1]=a[n];
	ForD(i,n-1) {
		int &k=h[i][0];
		val[i][++k]=a[i];h[i][k]=i;
		For(j,h[i+1][0]) {
			ll p=gcd(val[i][k],val[i+1][j]);
			if (p!=val[i][k]) 
				val[i][++k]=p;
			h[i][k]=h[i+1][j];
		}
	}
}

ll modify(){return 0;
}
ll modify(int l,int r,int f)
{
	ll ret=0;
	if (f==0) {  //left 
		int fro=l;
		For(j,h2[l][0]) {
			int last=min(h2[l][j],r);
			if (fro<=last) ret+=val2[l][j]*(last-fro+1);
			fro=last+1;
			if (fro>r) break;
		}
	} else {
		int last=r;
		For(j,h[r][0]) {
			int fro=max(h[r][j],l);
			if (fro<=last) ret+=val[r][j]*(last-fro+1);
			last=fro-1;
			if (fro<l) break;
		}
		
	}
	return ret;
	
}

int main()
{
//	freopen("B.in","r",stdin);
	
	int T; cin>>T;
	while(T--) {
		cin>>n;
		For(i,n) scanf("%d",&a[i]);
		init(h,val);init2(h2,val2);

		cin>>Q;
		MEM(ans) 
		For(i,Q)
		{
			 scanf("%d%d",&comm[i].l,&comm[i].r),comm[i].i=i;
		}
		sort(comm+1,comm+1+Q);
		
		int nowl=1,nowr=1;
		ll nowans=a[1];
		For(i,Q)
		{
			while (nowl<comm[i].l) nowans-=modify(nowl,nowr,0),nowl++;
			while (nowl>comm[i].l) nowans+=modify(nowl-1,nowr,0),nowl--;
			while (comm[i].r<nowr) nowans-=modify(nowl,nowr,1),nowr--;
			while (comm[i].r>nowr) nowans+=modify(nowl,nowr+1,1),nowr++;
			
			ans[comm[i].i]=nowans; 
		}
		
		For(i,Q) printf("%I64d\n",ans[i]);
	} 	
	
	return 0;
}