HDU 5869 Different GCD Subarray Query (GCD种类预处理+树状数组维护)

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

问你l~r之间的连续序列的gcd种类。

首先固定右端点，预处理gcd不同尽量靠右的位置(此时gcd种类不超过loga[i]种)。

预处理gcd如下代码，感觉真的有点巧妙...

         for(int i = ; i <= n; ++i) {
             int x = a[i], y = i;
             for(int j = ; j < ans[i - ].size(); ++j) {
                 int gcd = GCD(x, ans[i - ][j].first);
                 if(gcd != x) {
                     ans[i].push_back(make_pair(x, y));
                     x = gcd, y = ans[i - ][j].second;
                 }
             }
             ans[i].push_back(make_pair(x, y));
         }

然后用树状数组维护右端点固定的gcd位置。

 //#pragma comment(linker, "/STACK:102400000, 102400000")
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <vector>
 #include <cmath>
 #include <ctime>
 #include <list>
 #include <set>
 #include <map>
 using namespace std;
 typedef long long LL;
 typedef pair <int, int> P;
 const int N = 1e6 + ;
 int a[N/ + ], bit[N], _pos[N]; //_pos存的是gcd上一次出现的位置
 vector <P> ans[N/ + ]; //存的是以i为右端点的gcd
 struct query {
     int l, r, pos;
     bool operator <(const query& cmp) const {
         return r < cmp.r;
     }
 }q[N/ + ];
 int res[N/ + ]; //答案

 void init(int n) {
     memset(_pos, , sizeof(_pos));
     memset(bit, , sizeof(bit));
     for(int i = ; i <= n; ++i) {
         ans[i].clear();
     }
 }

 int GCD(int a, int b) {
     return b ? GCD(b, a % b): a;
 }

 void add(int i, int x) {
     for( ; i <= N; i += (i&-i))
         bit[i] += x;
 }

 int sum(int i) {
     int s = ;
     for( ; i >= ; i -= (i&-i))
         s += bit[i];
     return s;
 }

 int main()
 {
     int n, m;
     while(scanf("%d %d", &n, &m) != EOF) {
         init(n);
         for(int i = ; i <= n; ++i) {
             scanf("%d", a + i);
         }
         for(int i = ; i <= m; ++i) {
             scanf("%d %d", &q[i].l, &q[i].r);
             q[i].pos = i;
         }
         for(int i = ; i <= n; ++i) {
             int x = a[i], y = i;
             for(int j = ; j < ans[i - ].size(); ++j) {
                 int gcd = GCD(x, ans[i - ][j].first);
                 if(gcd != x) {
                     ans[i].push_back(make_pair(x, y));
                     x = gcd, y = ans[i - ][j].second;
                 }
             }
             ans[i].push_back(make_pair(x, y));
         }
         sort(q + , q + m + );
         int p = ;
         for(int i = ; i <= n; ++i) {
             for(int j = ; j < ans[i].size(); ++j) {
                 if(!_pos[ans[i][j].first]) {
                     add(ans[i][j].second, );
                     _pos[ans[i][j].first] = ans[i][j].second;
                 } else {
                     add(_pos[ans[i][j].first], -);
                     _pos[ans[i][j].first] = ans[i][j].second;
                     add(ans[i][j].second, );
                 }
             }
             while(i == q[p].r && p <= m) {
                 res[q[p].pos] = sum(q[p].r) - sum(q[p].l - );
                 ++p;
             }
         }
         for(int i = ; i <= m; ++i) {
             printf("%d\n", res[i]);
         }
     }
     return ;
 }