Jzzhu and Apples CodeForces - 449C (构造,数学)

大意: 求从[1,n]范围选择尽量多的数对, 使得每对数的gcd>1

考虑所有除2以外且不超过n/2的素数p, 若p倍数可以选择的有偶数个, 直接全部划分即可

有奇数个的话, 余下一个2*p不划分, 其余全部划分

最后再将2的倍数全部划分一下即可

#include <iostream>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <cstdio>
#include <vector>
#define x first
#define y second
#define pb push_back
#define REP(i,a,n) for(int i=a;i<=n;++i)
using namespace std;
typedef pair<int,int> pii;

const int N = 1e5+10;
int n;
int vis[N], p[N];
vector<pii> ans;

void seive(int n) {
	int mx = sqrt(n+0.5);
	REP(i,2,mx) if (!vis[i]) {
		for (int j=i*i; j<=n; j+=i) {
			vis[j] = 1;
		}
	}
	REP(i,3,n) if (!vis[i]) p[++*p]=i;
}

int main() {
	scanf("%d", &n);
	seive(n/2);
	memset(vis, 0, sizeof vis);
	vector<int> ret;
	REP(i,1,*p) {
		vector<int> num;
		for (int t=p[i]; t<=n; t+=p[i]) {
			if (!vis[t]) num.pb(t);
		}
		while (num.size()>=2) {
			int x = num.back();
			num.pop_back();
			vis[x] = 1;
			int y = num.back();
			num.pop_back();
			vis[y] = 1;
			if (y==2*p[i]) {
				ret.pb(y);
				y = p[i];
				vis[y] = 1;
			}
			ans.pb(pii(x,y));
		}
	}
	for (int t=2; t<=n; t<<=1) ret.pb(t);
	while (ret.size()>=2) {
		int x = ret.back();
		ret.pop_back();
		int y = ret.back();
		ret.pop_back();
		ans.pb(pii(x,y));
	}
	printf("%d\n", int(ans.size()));
	for (auto t:ans) {
		printf("%d %d\n",t.x,t.y);
	}
}