Codeforces 828D

High Load

题意：构造一颗n个节点k个叶子节点是树，使得叶子节点的最远距离最小

思路：贪心构造，先构造一个k+1个节点，k个叶子节点的树，用数组标记每个叶子节点的编号，然后剩下n-k-1个节点依次往每个叶子节点上加，并更新叶子节点

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const int N=2e5+100;

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int n,k; cin>>n>>k;
    int t=(n-1)/k, ans=t,x[N];
    if(n-(t*k+1)==0) ans+=t;
    else if(n-(t*k+1)==1) ans+=t+1;
    else ans+=t+2;
    cout<<ans<<endl;
    for(int i=1; i<=k; ++i){
        cout<<i<<" "<<k+1<<endl;
        x[i]=i;
    }
    int now=0;
    for(int i=k+2; i<=n; ++i){
        now%=k; 
        cout<<x[now+1]<<" "<<i<<endl;
        x[++now]=i;
    }
    return 0;
}