题目链接：hdu 4777 Rabbit Kingdom
题目大意：一个兔子王国，有N只兔子，每只兔子有一个重量，如果两只兔子的重量不互质，那么就会干架，现在国王想将l r之间的兔子关进监狱，它想知道会有多少只兔子不会和别的兔子干架。
解题思路：预处理出每只兔子的L，R表示向左和向右最近会与该兔子发生冲突的兔子，预处理的时候只要将每只兔子的重量分解成质因子后遍历两遍。
对于询问，将询问按照右区间排序，碰到i，则L位置+1，碰到R，则i位置+1，L位置-1。（如果L ≤ l && r ≤ R，那么兔子在这个l，r询问中是不会与其他兔子冲突）
这样l~r区间统计的即为会打架的兔子。
#include
#include
#include
#include
#include
using namespace std; const int maxn = 200005; int np, pri[maxn], vis[maxn]; void prime_table(int n) { np = 0; for (int i = 2; i <= n; i++) { if (vis[i]) continue; pri[np++] = i; for (int j = i * 2; j <= n; j += i) vis[j] = 1; } vis[1] = 1; } #define lowbit(x) ((x)&(-x)) struct Seg { int l, r, id; Seg (int l = 0, int r = 0, int id = 0) { this->l = l; this->r = r; this->id = id; } friend bool operator < (const Seg& a, const Seg& b) { return a.r < b.r; } }; int N, M, L[maxn], P[maxn]; int fenw[maxn], ans[maxn]; vector
g[maxn]; vector
vec, que; inline void add (int x, int d) { if (x <= 0) return; while (x <= N) { fenw[x] += d; x += lowbit(x); } } inline int sum (int x) { int ret = 0; while (x) { ret += fenw[x]; x -= lowbit(x); } return ret; } void init () { vec.clear(); que.clear(); int x; for (int i = 1; i <= N; i++) { scanf("%d", &x); g[i].clear(); for (int j = 0; j < np; j++) { if (x % pri[j] == 0) { g[i].push_back(pri[j]); while (x % pri[j] == 0) x /= pri[j]; } if (vis[x] == 0) { g[i].push_back(x); break; } } } for (int i = 0; i < maxn; i++) P[i] = 0; for (int i = 1; i <= N; i++) { int tmp = 0; for (int j = 0; j < g[i].size(); j++) { tmp = max(tmp, P[g[i][j]]); P[g[i][j]] = i; } L[i] = tmp; vec.push_back(Seg(tmp, i, 0)); } for (int i = 0; i < maxn; i++) P[i] = N + 1; for (int i = N; i; i--) { int tmp = N + 1; for (int j = 0; j < g[i].size(); j++) { tmp = min(tmp, P[g[i][j]]); P[g[i][j]] = i; } vec.push_back(Seg(i, tmp, 1)); } int l, r; for (int i = 1; i <= M; i++) { scanf("%d%d", &l, &r); que.push_back(Seg(l, r, i)); } } void solve () { sort(que.begin(), que.end()); sort(vec.begin(), vec.end()); memset(fenw, 0, sizeof(fenw)); int mv = 0; for (int i = 0; i < que.size(); i++) { while (mv < vec.size() && vec[mv].r <= que[i].r) { add(vec[mv].l, 1); if (vec[mv].id) add(L[vec[mv].l], -1); mv++; } int tmp = sum(que[i].r) - sum(que[i].l - 1); ans[que[i].id] = que[i].r - que[i].l + 1 - tmp; } } int main () { prime_table(maxn); while (scanf("%d%d", &N, &M) == 2 && N + M) { init(); solve(); for (int i = 1; i <= M; i++) printf("%d\n", ans[i]); } return 0; }