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2016大连网络赛 Different GCD Subarray Query

2024-08-12 22:51:40   224人阅读

Different GCD Subarray Query

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description
This is a simple problem. The teacher gives Bob a list of problems about GCD (Greatest Common Divisor). After studying some of them, Bob thinks that GCD is so interesting. One day, he comes up with a new problem about GCD. Easy as it looks, Bob cannot figure it out himself. Now he turns to you for help, and here is the problem:
  
  Given an array a of N positive integers a1,a2,?aN−1,aN; a subarray of a is defined as a continuous interval between a1 and aN. In other words, ai,ai+1,?,aj−1,aj is a subarray of a, for 1≤i≤j≤N. For a query in the form (L,R), tell the number of different GCDs contributed by all subarrays of the interval [L,R].
  
 

 

Input
There are several tests, process till the end of input.
  
  For each test, the first line consists of two integers N and Q, denoting the length of the array and the number of queries, respectively. N positive integers are listed in the second line, followed by Q lines each containing two integers L,R for a query.

You can assume that
  
    1≤N,Q≤100000
    
   1≤ai≤1000000
 

 

Output
For each query, output the answer in one line.
 

 

Sample Input
5 31 3 4 6 93 52 51 5
 

 

Sample Output
666
 
分析：对于每个点来说，向前的gcd很少，可以预处理出gcd出现的最后的左端点；
　　　然后离线查询右端点，每走到一个点，更新gcd出现的左端点，然后树状数组询问答案即可；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=1e5+10;using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}int n,m,k,t,a[maxn],c[maxn],pr[maxn*10],ans[maxn];vector<pii >b[maxn];vector<pii >q[maxn];void add(int x,int y){    for(int i=x;i<=n;i+=(i&(-i)))        c[i]+=y;}int get(int x){    int ret=0;    for(int i=x;i;i-=i&(-i))        ret+=c[i];    return ret;}void init(){    for(int i=1;i<=n;i++)    {        int x=a[i],y=i;        b[i].pb(mp(x,y));        for(int j=0;j<b[i-1].size();j++)        {            int now=gcd(b[i-1][j].fi,a[i]),pos=b[i-1][j].se;            if(now!=x)            {                b[i].pb(mp(now,pos));                x=now;            }        }    }}int main(){    int i,j;    while(~scanf("%d%d",&n,&m))    {        rep(i,1,n)scanf("%d",&a[i]),b[i].clear(),q[i].clear();        memset(c,0,sizeof c);        memset(pr,0,sizeof pr);        init();        rep(i,1,m)        {            int x,y;            scanf("%d%d",&x,&y);            q[y].pb(mp(x,i));        }        rep(i,1,n)        {            for(j=0;j<b[i].size();j++)            {                int x=b[i][j].fi,pos=b[i][j].se;                if(pr[x])add(pr[x],-1);                add(pos,1);                pr[x]=pos;            }            for(int j=0;j<q[i].size();j++)            {                int x=q[i][j].fi,y=q[i][j].se;                ans[y]=get(i)-(x-1==0?0:get(x-1));            }        }        rep(i,1,m)printf("%d\n",ans[i]);    }    //system("Pause");    return 0;}

2016大连网络赛 Different GCD Subarray Query


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