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...
1 for(int i = 1; i <= n; ++i) { 2 int x = a[i], y = i; 3 for(int j = 0; j < ans[i - 1].size(); ++j) { 4 int gcd = GCD(x, ans[i - 1][j].first); 5 if(gcd != x) { 6 ans[i].push_back(make_pair(x, y)); 7 x = gcd, y = ans[i - 1][j].second; 8 } 9 } 10 ans[i].push_back(make_pair(x, y)); 11 }
然后用树状数组维护右端点固定的gcd位置。
1 //#pragma comment(linker, "/STACK:102400000, 102400000") 2 #include <algorithm> 3 #include <iostream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <vector> 8 #include <cmath> 9 #include <ctime> 10 #include <list> 11 #include <set> 12 #include <map> 13 using namespace std; 14 typedef long long LL; 15 typedef pair <int, int> P; 16 const int N = 1e6 + 5; 17 int a[N/10 + 5], bit[N], _pos[N]; //_pos存的是gcd上一次出现的位置 18 vector <P> ans[N/10 + 5]; //存的是以i为右端点的gcd 19 struct query { 20 int l, r, pos; 21 bool operator <(const query& cmp) const { 22 return r < cmp.r; 23 } 24 }q[N/10 + 5]; 25 int res[N/10 + 5]; //答案 26 27 void init(int n) { 28 memset(_pos, 0, sizeof(_pos)); 29 memset(bit, 0, sizeof(bit)); 30 for(int i = 1; i <= n; ++i) { 31 ans[i].clear(); 32 } 33 } 34 35 int GCD(int a, int b) { 36 return b ? GCD(b, a % b): a; 37 } 38 39 void add(int i, int x) { 40 for( ; i <= N; i += (i&-i)) 41 bit[i] += x; 42 } 43 44 int sum(int i) { 45 int s = 0; 46 for( ; i >= 1; i -= (i&-i)) 47 s += bit[i]; 48 return s; 49 } 50 51 int main() 52 { 53 int n, m; 54 while(scanf("%d %d", &n, &m) != EOF) { 55 init(n); 56 for(int i = 1; i <= n; ++i) { 57 scanf("%d", a + i); 58 } 59 for(int i = 1; i <= m; ++i) { 60 scanf("%d %d", &q[i].l, &q[i].r); 61 q[i].pos = i; 62 } 63 for(int i = 1; i <= n; ++i) { 64 int x = a[i], y = i; 65 for(int j = 0; j < ans[i - 1].size(); ++j) { 66 int gcd = GCD(x, ans[i - 1][j].first); 67 if(gcd != x) { 68 ans[i].push_back(make_pair(x, y)); 69 x = gcd, y = ans[i - 1][j].second; 70 } 71 } 72 ans[i].push_back(make_pair(x, y)); 73 } 74 sort(q + 1, q + m + 1); 75 int p = 1; 76 for(int i = 1; i <= n; ++i) { 77 for(int j = 0; j < ans[i].size(); ++j) { 78 if(!_pos[ans[i][j].first]) { 79 add(ans[i][j].second, 1); 80 _pos[ans[i][j].first] = ans[i][j].second; 81 } else { 82 add(_pos[ans[i][j].first], -1); 83 _pos[ans[i][j].first] = ans[i][j].second; 84 add(ans[i][j].second, 1); 85 } 86 } 87 while(i == q[p].r && p <= m) { 88 res[q[p].pos] = sum(q[p].r) - sum(q[p].l - 1); 89 ++p; 90 } 91 } 92 for(int i = 1; i <= m; ++i) { 93 printf("%d\n", res[i]); 94 } 95 } 96 return 0; 97 }
