传送门
题意：从1-n能配多少对 公约数不为1的对
#include <bits/stdc++.h> //#include <ext/pb_ds/tree_policy.hpp> //#include <ext/pb_ds/assoc_container.hpp> //using namespace __gnu_pbds; using namespace std; #define pi acos(-1) #define endl '\n' #define me(x) memset(x,0,sizeof(x)); #define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++) #define close() ios::sync_with_stdio(0); #define rand() srand(time(0)); typedef long long LL; const int INF=0x3f3f3f3f; const LL LINF=0x3f3f3f3f3f3f3f3fLL; //const int dx[]={-1,0,1,0,-1,-1,1,1}; //const int dy[]={0,1,0,-1,1,-1,1,-1}; const int maxn=1e3+5; const int maxx=1e5+100; const double EPS=1e-9; const int MOD=1000000007; #define mod(x) ((x)%MOD); template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);} template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);} template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));} template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));} //typedef tree<pt,null_type,less< pt >,rb_tree_tag,tree_order_statistics_node_update> rbtree; /*lch[root] = build(L1,p-1,L2+1,L2+cnt); rch[root] = build(p+1,R1,L2+cnt+1,R2);中前*/ /*lch[root] = build(L1,p-1,L2,L2+cnt-1); rch[root] = build(p+1,R1,L2+cnt,R2-1);中后*/ long long gcd(long long a , long long b){if(b==0) return a;a%=b;return gcd(b,a);} inline int Scan() { int res=0,ch,flag=0; if((ch=getchar())=='-')flag=1; else if(ch>='0' && ch<='9')res=ch-'0'; while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0'; return flag ? -res : res; } typedef pair<int, int> pii; bool iscomp[maxx+5], vis[maxx+5]; void prime_table(int n) //素数打表 { for (int i = 2; i * i <= n; i++) { if (iscomp[i]) continue; for (int j = i * i; j <= n; j += i) iscomp[j] = 1; } } int main () { int n; scanf("%d", &n); prime_table(n); vector<int> g; vector<pii> ans; for (int i = n / 2; i > 1; i--)//从n/2到1枚举 { if (iscomp[i])//合数就跳过 continue; g.clear(); for (int j = i; j <= n; j += i) { if (vis[j] == 0) g.push_back(j); } if (g.size() & 1) swap(g[1], g[g.size()-1]);//拿出2pi与其他进行匹配 for (int i = 0; i < g.size() - 1; i += 2) { ans.push_back(make_pair(g[i], g[i+1])); vis[g[i]] = vis[g[i+1]] = 1; } } printf("%lu\n", ans.size()); for (int i = 0; i < ans.size(); i++) printf("%d %d\n", ans[i].first, ans[i].second); return 0; }
