HDU 4913(Least common multiple-线段树+容斥)

Least common multiple

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 767    Accepted Submission(s): 257

Problem Description

bobo has an integer set S={x ₁,x ₂,…,x _n}, where x _i=2 ^ai * 3 ^bi.

For each non-empty subsets of S, bobo added the LCM (least common multiple) of the subset up. Find the sum of LCM modulo (10 ⁹+7).

 

Input

The input consists of several tests. For each tests:

The first line contains n (1≤n≤10 ⁵). Each of the following n lines contain 2 integers a _i,b _i (0≤a _i,b _i≤10 ⁹).

 

Output

For each tests:

A single integer, the value of the sum.

 

Sample Input

   
   
   
   

    
    
    
    2
0 1
1 0
3
1 2
2 1
1 2
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    11
174
   
   
   
   

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

Recommend

We have carefully selected several similar problems for you:   5426  5425  5424  5423  5422 

 

按Ai从小到大，Bi从小到大加入，考虑加入的数对答案的影响

b<Bi有cnt个 故 

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<vector>
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 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 (o<<1)
#define Rson ((o<<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 MEM2(a,i) memset(a,i,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (100000+10)
#define fi first
#define se second
#define mp make_pair
typedef __int64 ll;
ll mul(ll a,ll b){return (a%F*b%F)%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;}
ll pow2(ll a,ll b)
{
	if (b==0) return 1;
	if (b==1) return a;
	ll p = pow2(a,b/2);
	p=p*p%F;
	if ( b & 1 ) p = p * a%F;
	return p;
} 

class SegmentTree  
{  
    ll a[MAXN*4],sumv[MAXN*4],a2[MAXN*4],sumv2[MAXN*4];  
    int mark[MAXN*4];
    int n;  
public:  
    SegmentTree(){MEM(a) MEM(sumv) MEM(a2) MEM(sumv2) MEM(mark) }  
    SegmentTree(int _n):n(_n){MEM(a) MEM(sumv) MEM(a2) MEM(sumv2) MEM(mark)  }  
	void mem(int _n)  
    {  
        n=_n;  
        MEM(a) MEM(sumv) MEM(a2) MEM(sumv2) MEM(mark) 
    }  
    void maintain(int o,int L,int R)  
    {
    	if (L<R) 
		{
			sumv[o]=(sumv[Lson]+sumv[Rson])%F;
			sumv2[o]=(sumv2[Lson]+sumv2[Rson])%F;

		}  
		
    }

	int y1,y2;
	ll v;
	void update(int o,int L,int R) //y1,y2,v
	{
		if (L==R) {
			sumv[o]++; 
			sumv2[o]=(sumv2[o]+v)%F;
			return ;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update(Lson,L,M); 
			else update(Rson,M+1,R); 
		}
		maintain(o,L,R); 
	}
	void update2(int o,int L,int R) 
	{
		if (y1<=L&&R<=y2) {
			mark[o]++;sumv2[o]=(2*sumv2[o])%F;
			return;
		}
		else{
			pushdown(o);
			int M=(R+L)>>1;
			if (y1<=M) update2(Lson,L,M); 
			if (M< y2) update2(Rson,M+1,R); 
		}
		maintain(o,L,R); 
	}
	
	void pushdown(int o) 
	{ 
		if (mark[o])
		{ 
			mark[Lson]+=mark[o];
			mark[Rson]+=mark[o];
			ll t=pow2(2,mark[o]);
			sumv2[Lson]=sumv2[Lson]*t%F;
			sumv2[Rson]=sumv2[Rson]*t%F;
			mark[o]=0;
		}
	}
	ll _sum,_sum2; 
	
	void query2(int o,int L,int R)
	{
		if (y1<=L&&R<=y2)
		{
			upd(_sum,sumv[o]);
			upd(_sum2,sumv2[o]);
			return;
		} else {
			pushdown(o);
			int M=(L+R)>>1;
			if (y1<=M) query2(Lson,L,M);
			if (M< y2) query2(Rson,M+1,R);
		}
	}
	void add(int l,ll v)
	{
		y1=y2=l;this->v=v;
		update(1,1,n);
	}
	void mul(int l,int r)
	{
		if (l>r) return ; 
		y1=l,y2=r;
		update2(1,1,n);
	}
	ll ask(int l,int r,int b=1)
	{
		if (l>r) return 0; 
		_sum=_sum2=0;
		y1=l,y2=r;
		query2(1,1,n);
		if (b==1) return _sum;
		return _sum2;		
	}
	void print()
	{
		For(i,n)
			cout<<ask(i,i,2)<<' ';
		cout<<endl;
	}

}S;  

int n;
pair<ll,ll> p[MAXN];
ll bb[MAXN];
int main()
{
//	freopen("hdu4913.in","r",stdin);
//	freopen(".out","w",stdout);
	
	while(scanf("%d",&n)==1) {
		int a,b;
		For(i,n) scanf("%d%d",&a,&b),p[i]=mp(a,b),bb[i]=p[i].se;
		sort(p+1,p+1+n);
		
		sort(bb+1,bb+1+n);
		int m=unique(bb+1,bb+1+n)-(bb+1);
		
		S.mem(m); 
		ll ans = 0;
		
		For(i,n) {
			ll a=p[i].fi,b=p[i].se;
			ll now=pow2(2LL,a),now3=pow2(3LL,b);
			ll nowv=mul(now,now3);
			int pos = lower_bound( bb+1 , bb+1+m , p[i].se ) - (bb);
			
			ll cnt=S.ask( 1,pos,1);
			ll tmp=pow2(2,cnt);

			upd( ans , nowv * tmp % F );
			upd( ans , now * S.ask(pos+1,m,2) % F);
			S.mul(pos+1,m);
			
			S.add(pos,mul(now3,tmp));
			
		} 
		printf("%I64d\n",ans%F); 
	}
	
	return 0;
}