hdu 5869 区间gcd的求法及应用

#include <stdio.h>  
#include <string.h>  
#include <iostream>  
#include <algorithm>  
#include <stack>  
#include <vector>  
#include <queue>  
#include <set>  
#include <map>  
#include <string>  
#include <math.h>  
#include <stdlib.h>  
#include <time.h> 
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#define PF(x) cout << "debug: " << x << " ";
#define EL cout << endl;
#define PC(x) puts(x);
typedef long long ll;
#define CLR(x, v) sizeof (x, v, sizeof(x))
using namespace std;
const int INF = 0x5f5f5f5f;
const int maxn = 2e5 + 10;
const int mod = 1e9 + 7;
const double eps = acos(-1);
const double PI= atan(1.0)*4;
int n,q,a[maxn],tree[maxn],vis[2000000],ans[maxn];
vector<pair<int,int> >gd[maxn];
struct st{
    int l,r,id;
}qer[maxn];
int gcd(int a,int b){
    return b == 0?a:gcd(b,a%b);
}
bool cmp(st x,st y){
    return x.r < y.r;
}
void Add(int k,int num){
    while(k <= n){
        tree[k] += num;
        k += k&(-k);
    }
}
int Sum(int k){
    int sum = 0;
    while(k > 0){
        sum += tree[k];
        k -= k&(-k);
    }
    return sum;
}
int main(){
    freopen("in.txt","r",stdin);
    while(~scanf("%d%d",&n,&q)){
        for(int i = 1;i <= n;i++){
            scanf("%d",&a[i]);
            gd[i].clear();
        }
        for(int i = 1;i <= n;i++){
            int x = a[i],y = i;
            for(int j = 0;j <gd[i-1].size();j++){
                int t = gcd(gd[i-1][j].first,a[i]);
                if(t != x){
                    gd[i].push_back(make_pair(x,y));
                    x = t,y = gd[i-1][j].second;//x记录从y位置到当前处理的i位置的gcd值，y为相同gcd中最右边的位置，最右边这个是必须的
                }
            }
            gd[i].push_back(make_pair(x,y));
        }
       for(int i = 1;i <= q;i++){
            scanf("%d%d",&qer[i].l,&qer[i].r);
            qer[i].id = i;
       }
        sort(qer + 1,qer + 1 + q,cmp);
        memset(vis,0,sizeof(vis));
        memset(tree,0,sizeof(tree));
        int len = 1;
        for(int i = 1; i <= n;i++){
           // cout<<i<<":";
            for(int j = 0;j < gd[i].size();j++){
               // cout<<gd[i][j].first<<"->"<<gd[i][j].second<<" ";
                int c1 = gd[i][j].first,c2 = gd[i][j].second;
              //如果这个gcd值c1出现过则更新其位置，根据非严格递减，                             这样一定能更新为最靠近当前的i的位置
                if(vis[c1] > 0)                  
                    Add(vis[c1],-1);
                vis[c1] = c2;
                Add(c2,1);
            }
           // cout<<endl;
            while(qer[len].r == i){
                ans[qer[len].id] = Sum(qer[len].r) - Sum(qer[len].l-1);
                len++;
            }
        }
        for(int i = 1;i <= q;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}