[HDOJ5869] Different GCD Subarray Query（RMQ，树状数组，离线）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5869

题意：n个数，q次询问，问区间内gcd不同值的个数。

和dquery那道题一样，也是离线的做法。按照查询的r从小到大排序，每插入一个数字ai，则更新一次gcd，总是把gcd出现向后移动，这样可以满足靠后的查询能够包含住这些值。

首先用st表预处理出区间的gcd，而且观察到gcd单调不增的性质，在枚举区间gcd的值的时候，可以二分。

#include <bits/stdc++.h>
using namespace std;

typedef struct Event {
    int l, r, id;
}Event;
const int maxn = 100100;
int n, q, a[maxn];
int dp[maxn][20];
int bit[maxn];
vector<Event> event;
int ret[maxn];
unordered_map<int, int> vis;

bool cmp(Event a, Event b) { return a.r < b.r; }
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
int lowbit(int x) { return x & (-x); }
void add(int i, int v) { for(; i <= n; i+=lowbit(i)) bit[i] += v; }
int sum(int i) { int ret = 0; for(; i > 0; i-=lowbit(i)) ret += bit[i]; return ret; }
int query(int l, int r) { int k = int(log(r-l+1.0)/log(2.0)); return gcd(dp[l][k], dp[r-(1<<k)+1][k]); }

int main() {
    // freopen("in", "r", stdin);
    int l, r;
    while(~scanf("%d%d",&n,&q)) {
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        event.clear(); vis.clear();
        memset(bit, 0, sizeof(bit));
        for(int i = 1; i <= n; i++) dp[i][0] = a[i];
        int k = int(log(n+1.0)/log(2.0));
        for(int j = 1; j <= k; j++) {
            for(int i = 1; i + (1 << j) - 1 <= n; i++) {
                dp[i][j] = gcd(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
            }
        }
        for(int i = 1; i <= q; i++) {
            scanf("%d%d",&l,&r);
            event.push_back(Event{l,r,i});
        }
        sort(event.begin(), event.end(), cmp);
        int pos = 0;
        for(int i = 1; i <= n; i++) {
            int cur = a[i];
            int j = i;
            while(j >= 1) {
                int tmp = j;
                l = 1, r = j;
                while(l <= r) {
                    int mid = (l + r) >> 1;
                    if(l == r && query(mid, i) == cur) {
                        tmp = mid;
                        break;
                    }
                    if(query(mid, i) == cur) {
                        tmp = mid;
                        r = mid - 1;
                    }
                    else l = mid + 1;
                }
                if(vis.find(cur) == vis.end()) {
                    add(j, 1);
                    vis[cur] = j;
                }
                else if(vis[cur] < j) {
                    add(vis[cur], -1);
                    add(j, 1);
                    vis[cur] = j;
                }
                j = tmp - 1;
                if(j >= 1) cur = query(j ,i);
            }
            while(pos < event.size() && event[pos].r == i) {
                ret[event[pos].id] = sum(event[pos].r) - sum(event[pos].l-1);
                pos++;
            }
        }
        for(int i = 1; i <= q; i++) {
            printf("%d
", ret[i]);
        }
    }
    return 0;
}