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={x₁,x₂,…,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 (10⁹+7).

 

Input

The input consists of several tests. For each tests:

The first line contains n (1≤n≤10⁵). Each of the following n lines contain 2 integers a_i,b_i (0≤a_i,b_i≤10⁹).

 

Output

For each tests:

A single integer, the value of the sum.

 

Sample Input

20 11 031 22 11 2

 

Sample Output

11174

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

 

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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_pairtypedef __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;}