FZU 2224 An exciting GCD problem(GCD种类预处理+树状数组维护)同hdu5869
题目链接：http://acm.fzu.edu.cn/problem.php?pid=2224
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 map <int, int> _pos; 20 struct query { 21 int l, r, pos; 22 bool operator <(const query& cmp) const { 23 return r < cmp.r; 24 } 25 }q[N/10 + 5]; 26 int res[N/10 + 5]; //答案 27 28 void init(int n) { 29 _pos.clear(); 30 memset(bit, 0, sizeof(bit)); 31 for(int i = 1; i <= n; ++i) { 32 ans[i].clear(); 33 } 34 } 35 36 int GCD(int a, int b) { 37 return b ? GCD(b, a % b): a; 38 } 39 40 void add(int i, int x) { 41 for( ; i <= N; i += (i&-i)) 42 bit[i] += x; 43 } 44 45 int sum(int i) { 46 int s = 0; 47 for( ; i >= 1; i -= (i&-i)) 48 s += bit[i]; 49 return s; 50 } 51 52 int main() 53 { 54 int n, m, t; 55 scanf("%d", &t); 56 while(t--) { 57 scanf("%d %d", &n, &m); 58 init(n); 59 for(int i = 1; i <= n; ++i) { 60 scanf("%d", a + i); 61 } 62 for(int i = 1; i <= m; ++i) { 63 scanf("%d %d", &q[i].l, &q[i].r); 64 q[i].pos = i; 65 } 66 for(int i = 1; i <= n; ++i) { 67 int x = a[i], y = i; 68 for(int j = 0; j < ans[i - 1].size(); ++j) { 69 int gcd = GCD(x, ans[i - 1][j].first); 70 if(gcd != x) { 71 ans[i].push_back(make_pair(x, y)); 72 x = gcd, y = ans[i - 1][j].second; 73 } 74 } 75 ans[i].push_back(make_pair(x, y)); 76 } 77 sort(q + 1, q + m + 1); 78 int p = 1; 79 for(int i = 1; i <= n; ++i) { 80 for(int j = 0; j < ans[i].size(); ++j) { 81 if(!_pos[ans[i][j].first]) { 82 add(ans[i][j].second, 1); 83 _pos[ans[i][j].first] = ans[i][j].second; 84 } else { 85 add(_pos[ans[i][j].first], -1); 86 _pos[ans[i][j].first] = ans[i][j].second; 87 add(ans[i][j].second, 1); 88 } 89 } 90 while(i == q[p].r && p <= m) { 91 res[q[p].pos] = sum(q[p].r) - sum(q[p].l - 1); 92 ++p; 93 } 94 } 95 for(int i = 1; i <= m; ++i) { 96 printf("%d\n", res[i]); 97 } 98 } 99 return 0; 100 }
