BZOJ 2632: [neerc2011]Gcd guessing game

最坏情况下每次得到的答案都是 \(1\)
而每次猜一个数都相当于筛去一部分质数
那就是把 \(1\) 到 \(n\) 之间的质因数分组，每一组里乘积都小于等于 \(n\)
求最小组数
这个用双指针，为每个大质数分配若干个小质数即可

#include <bits/stdc++.h>
#define pb push_back
#define fi first
#define se second
#define pii pair<int, int>
#define pli pair<ll, int>
#define lp p << 1
#define rp p << 1 | 1
#define mid ((l + r) / 2)
#define lowbit(i) ((i) & (-i))
#define ll long long
#define ull unsigned long long
#define db double
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define Edg int ccnt=1,head[N],to[E],ne[E];void addd(int u,int v){to[++ccnt]=v;ne[ccnt]=head[u];head[u]=ccnt;}void add(int u,int v){addd(u,v);addd(v,u);}
#define Edgc int ccnt=1,head[N],to[E],ne[E],c[E];void addd(int u,int v,int w){to[++ccnt]=v;ne[ccnt]=head[u];c[ccnt]=w;head[u]=ccnt;}void add(int u,int v,int w){addd(u,v,w);addd(v,u,w);}
#define es(u,i,v) for(int i=head[u],v=to[i];i;i=ne[i],v=to[i])
const int MOD = 1e9 + 7;
void M(int &x) {if (x >= MOD)x -= MOD; if (x < 0)x += MOD;}
int qp(int a, int b = MOD - 2) {int ans = 1; for (; b; a = 1LL * a * a % MOD, b >>= 1)if (b & 1)ans = 1LL * ans * a % MOD; return ans % MOD;}
template<class T>T gcd(T a, T b) { while (b) { a %= b; std::swap(a, b); } return a; }
template<class T>bool chkmin(T &a, T b) { return a > b ? a = b, 1 : 0; }
template<class T>bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
char buf[1 << 21], *p1 = buf, *p2 = buf;
inline char getc() {
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
inline int _() {
	int x = 0, f = 1; char ch = getc();
	while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getc(); }
	while (ch >= '0' && ch <= '9') { x = x * 10ll + ch - 48; ch = getc(); }
	return x * f;
}

const int N = 1e4 + 7;
int n, prime[N], prin;
bool vis[N];
void init() {
	rep (i, 2, n + 1) {
		if (!vis[i]) prime[++prin] = i;
		rep (j, 1, prin + 1) {
			int v = i * prime[j];
			if (v > n) break;
			vis[v] = 1;
			if (i % prime[j] == 0) break;
		}
	}
}

int main() {
#ifdef LOCAL
	freopen("ans.out", "w", stdout);
#endif
	n = _();
	init();
	int l = 1, r = prin;
	int ans = 0;
	while (l <= r) {
		ans++;
		int x = prime[r];
		while (l < r && x * prime[l] <= n) x *= prime[l++];
		if (l == r) break;
		r--;
	}
	printf("%d\n", ans);
	return 0;
}