HDU 5869 Different GCD Subarray Query 离线+树状数组

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 ].
  

 

 

Input

 

There are several tests, process till the end of input.
  
  For each test, the first line consists of two integers 1000000

 

 

Output

 

For each query, output the answer in one line.

 

 

Sample Input

 

5 3 1 3 4 6 9 3 5 2 5 1 5

 

 

Sample Output

 

6 6 6

 

题意：

　　长度n的序列, m个询问区间[L, R], 问区间内的所有子段的不同GCD值有多少种.

题解：

　　固定右端点

　　预处理出 每个点向左延伸 的 不同gcd值

　　这样的 值不会超过log a 个

　　然后问题就变成了 问你一段区间内不同 gcd 值有多少，值是很少的 （询问一个区间有多少颜色的题型）

　　树状数组维护就可以了

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 1e6+10, M = 1e2+11, mod = 1e9+7, inf = 2e9;

int n,q,a[N],ans[N];
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b);}
vector<pii> G[N];
struct QQ{
        int l,r,id;
        bool operator < (const QQ &a) const {
             return a.r > r;
        }
}Q[N];
int C[N],vis[N];
void update(int x,int c) {
        for(int i =x; i < N; i+=i&(-i)) C[i] += c;
}
int ask(int x) {
        int s =0 ;
        for(int i = x; i; i-= i & (-i))s += C[i];
        return s;
}
int main() {
        while(scanf("%d%d",&n,&q)!=EOF) {
            for(int i = 1; i <= n; ++i) scanf("%d",&a[i]);
            for(int i = 0; i <= n; ++i) G[i].clear();
            for(int i = 1; i <= n; ++i) {
                    int x = a[i];
                    int y = i;
                    for(int j = 0; j < G[i-1].size(); ++j) {
                        int res = gcd(x,G[i-1][j].first);
                        if(x != res) {
                                G[i].push_back(MP(x,y));
                                x = res;
                                y = G[i-1][j].second;
                        }
                    }
                    G[i].push_back(MP(x,y));
            }
            memset(C,0,sizeof(C));
            memset(vis,0,sizeof(vis));
            for(int i = 1; i <= q; ++i) {scanf("%d%d",&Q[i].l,&Q[i].r),Q[i].id = i;}
            sort(Q+1,Q+q+1);
            for(int R = 0, i = 1; i <= q; ++i) {
                    while(R < Q[i].r) {
                        R++;
                        for(int j = 0; j < G[R].size(); ++j) {
                            int res = G[R][j].first;
                            int ids = G[R][j].second;
                            if(vis[res]) update(vis[res],-1);
                            vis[res] = ids;
                            update(vis[res],1);
                        }
                    }
                    ans[Q[i].id] = ask(R) - ask(Q[i].l-1);
            }
            for(int i = 1; i <= q; ++i) cout<<ans[i]<<endl;
        }
        return 0;
}