2016 Dalian Network Competition --- Different GCD Subarray Query (GCD discretization + tree array), tree array discretization

2016 Dalian Network Competition --- Different GCD Subarray Query (GCD discretization + tree array), tree array discretization

Question Link

Http://acm.split.hdu.edu.cn/showproblem.php? Pid = 1, 5869

 

Problem DescriptionThis 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, between aN −1, aN; a subarray of a is defined as a continuous interval between a1 and. in other words, ai, ai + 1, clerk, 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].
InputThere 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 OutputFor each query, output the answer in one line. Sample Input5 31 3 4 6 93 52 51 5 Sample Output666 Source2016 ACM/ICPC Asia Regional Dalian Online RecommendWange2014 | We have carefully selected several similar problems for you: 5877 5876 5874 5873 5872 question: Enter N and Q to indicate a sequence with N numbers, input l and r each time to represent an interval, and calculate the number of the maximum public multiples of the interval (obtained from the subinterval of this interval). Train of Thought: Perform GCD discrete processing on the series (~ I also know that such discretization still exists ~)

for(int i=1;i<=N;i++)        {            int tot=a[i],pos=i;            for(int j=0;j<v[i-1].size();j++)            {                int  r=__gcd(a[i],v[i-1][j].first);                if(tot!=r)                {                   v[i].push_back(make_pair(tot,pos));                   tot=r;  pos=v[i-1][j].second;                }            }            v[i].push_back(make_pair(tot,pos));        }

Then, the Q-requests are processed offline. the Q-requests are first input and sorted by the right endpoint from small to large. I ranges from 1 ~ N loop, when I = node [len]. r, ans [node [len]. id] = Sum (I)-Sum (node [len]. l-1). The Code is as follows:

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cmath>#include <map>#include <vector>using namespace std;int a[100005];int c[1000005];int vis[1000005];int sum[100005];struct Node{    int l,r;    int id;}node[100005];bool cmp(const Node s1,const Node s2){    return s1.r<s2.r;}vector<pair<int,int> > v[100005];int __gcd(int x,int y){    int r=x%y;    x=y;    y=r;    if(r==0) return x;    return __gcd(x,y);}int Lowbit(int t){    return t&(t^(t-1));}int Sum(int x){    int sum = 0;    while(x > 0)    {        sum += c[x];        x -= Lowbit(x);    }    return sum;}void add(int li,int t){    while(li<=1000005)    {        c[li]+=t;        li=li+Lowbit(li);    }}int main(){    int N,Q;    while(scanf("%d%d",&N,&Q)!=EOF)    {        for(int i=1;i<=N;i++) scanf("%d",&a[i]);        for(int i=1;i<=N;i++)        {            int tot=a[i],pos=i;            for(int j=0;j<v[i-1].size();j++)            {                int  r=__gcd(a[i],v[i-1][j].first);                if(tot!=r)                {                   v[i].push_back(make_pair(tot,pos));                   tot=r;  pos=v[i-1][j].second;                }            }            v[i].push_back(make_pair(tot,pos));        }        for(int i=0;i<Q;i++)            scanf("%d%d",&node[i].l,&node[i].r),node[i].id=i;        sort(node,node+Q,cmp);        memset(c,0,sizeof(c));        memset(vis,0,sizeof(vis));        int len=0;        for(int i=1;i<=N;i++)        {            for(int j=0;j<v[i].size();j++)            {                int s1=v[i][j].first;                int s2=v[i][j].second;                if(vis[s1]){                    add(vis[s1],-1);                }                vis[s1]=s2;                add(s2,1);            }            while(node[len].r==i)            {                sum[node[len].id]=Sum(i)-Sum(node[len].l-1);                len++;            }        }        for(int i=0;i<Q;i++)            printf("%d\n",sum[i]);        for(int i=0;i<=N;i++)            v[i].clear();    }    return 0;}