【Codeforces Round #547 (Div. 3)】 A B C D E F1 F2 G

其他 2019-03-23 00:06:49 阅读次数: 0

A

题意 问你从一个n 到 m 经过多少次数 *2 *3能到m

所以特判一下m能否整除n  然后除了以后根据奇偶去判断求次数

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;

int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    long long n,m;
    scanf("%lld%lld",&n,&m);
    if(m%n!=0)
    {
        printf("-1\n");
        return 0;
    }
    m/=n;
    int ans = 0;
    while(m%2==0)
    {
        ans++;
        m/=2;
    }
    while(m%3==0)
    {
        ans++;
        m/=3;
    }
    if(m==1)
    {
        printf("%d\n",ans);
    }
    else printf("-1\n");
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

B

给你一段 0 1 串 第一天接着第二天 问你最多 1 的次数 保证至少有个0

做法 答案为 max(中间最大连续1,后面连续1 + 前面连续1)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
int arr[200025];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    scanf("%d",&n);
    for(int i = 1;i<=n;++i)  scanf("%d",&arr[i]);
    int Back = 0,Front = 0,maxx = 0,res = 0;
    for(int i = 1;i<=n;++i)
    {
        if(arr[i]==1) Front++;
        else break;
    }
    for(int i = n;i>=1;i--)
    {
        if(arr[i]==1) Back++;
        else break;
    }
    for(int i= 1;i<=n;++i)
    {
        if(arr[i]==1) res++;
        else maxx = max(res,maxx),res=0;
    }
    maxx = max(res,maxx);
    printf("%d\n",max(maxx,Front+Back));
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

C

给你 n - 1 个差分数组 问你能否满足生成一个序列 1 - n

我们坐前缀和 就知道 第 i 个前缀和表示的是 arr[i] 与 a[1]的差 我们可以通过 1 - n的高斯公式求出 arr[1] 然后暴力判断一下即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 200025;
vector<int > vt,ans;
int q[MAX_N],cnt[MAX_N*3];
long long sum[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    scanf("%d",&n);
    bool flag = true;
    long long ck = 1ll*(n+1)*n/2,tmp = 0;
    for(int i = 1;i<n;++i)
    {
        scanf("%d",&q[i]);
        sum[i] = sum[i-1] + q[i];
        tmp+=sum[i];
    }
        long long dk = (ck - tmp)/n;
        if(dk*n!=ck-tmp)
        {
            printf("-1\n");
            return 0;
        }
        else
        {
            ans.push_back(dk);
            for(int i = 1;i<n;++i)
                ans.push_back(dk+sum[i]);
            sort(ans.begin(),ans.end());
            for(int i = 0;i<n;++i)
            {
                if(ans[i]!=(i+1))
                {
                    printf("-1\n");
                    return 0;
                }
            }
            printf("%lld ",dk);
            for(int i = 1;i<n;++i)
                printf("%lld ",dk+sum[i]);
        }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

D

题意 给你两个字符串 ?是可以随意匹配的 然后同字符可以匹配 问你能匹配的最长匹配 以及下标

做法 我们先把相同的预处理出来 然后处理第一个字符串剩余的 ? 然后处理第二个字符串剩余的 ? 即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
map<char,int >mp;
vector<pll> ans;
queue<int > q[28],q_[28],tmp,vt,vt_,st,st_;
bool flag[150025],flag_[150025];
char str[150025],str_[150025];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    for(int i = 0;i<26;++i)
        mp['a'+i] = i;
    mp['?'] = 26;
    int n,ANS = 0;
    scanf("%d",&n);
    scanf("%s",str+1);
    scanf("%s",str_+1);
    for(int i = 1;i<=n;++i)
    {
        if(str[i]=='?') vt.push(i);
        if(str_[i]=='?') vt_.push(i);
        else q[mp[str_[i]]].push(i);
    }
    for(int i = 1;i<=n;++i)
    {
        if(str[i]=='?') continue;
        else
        {
            if(!q[mp[str[i]]].empty())
            {
                int now = q[mp[str[i]]].front();
                ans.push_back(pll(i,now));
                flag[i] = true;
                flag_[now] = true;
                q[mp[str[i]]].pop();
            }
        }
    }
    for(int i = 1;i<=n;++i)
    {
        if(str[i]=='?'||flag[i]) continue;
        st.push(i);
    }
    for(int i = 1;i<=n;++i)
    {
        if(str_[i]=='?'||flag_[i]) continue;
        st_.push(i);
    }
    while(!vt.empty())
    {
        if(!st_.empty())
        {
            int now = vt.front();
            int now_ = st_.front();
            ans.push_back(pll(now,now_));
            vt.pop();
            st_.pop();
        }
        else if(!vt_.empty())
        {
            int now = vt.front();
            int now_ = vt_.front();
            ans.push_back(pll(now,now_));
            vt.pop();
            vt_.pop();
        }
    }
    while(!vt_.empty())
    {
        if(!st.empty())
        {
            int now_ = vt_.front();
            int now = st.front();
            ans.push_back(pll(now,now_));
            vt_.pop();
            st.pop();
        }
        else if(!vt_.empty())
        {
            int now = vt.front();
            int now_ = vt_.front();
            ans.push_back(pll(now,now_));
            vt.pop();
            vt_.pop();
        }
    }
    int sz = ans.size();
    printf("%d\n",sz);
    for(int i = 0;i<sz;++i)
        printf("%d %d\n",ans[i].first,ans[i].second);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

E

题意 给你 n 天的每天贡献 每天要使 H 加上贡献 问你最少天数达到 H <= 0

做法 我们先暴力枚举一遍 看第一次能不能打到H<=0 能就直接输出最小天数 然后处理一下轮次 暴力跑枚举

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 200025;
long long arr[MAX_N],sum_[MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    long long h,n,sum = 0,ans = 0;
    scanf("%lld%lld",&h,&n);
    for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]),sum+=arr[i],sum_[i] = sum_[i-1] + arr[i];
    long long h_ = h;
    for(int i = 1;i<=n;++i)
    {
        h_+=arr[i];
        ans++;
        if(h_<=0)
        {
            break;
        }
    }
    if(h_<=0)
    {
        printf("%lld\n",ans);
        return 0;
    }
    else if(sum>=0)
    {
        printf("-1\n");
        return 0;
    }

    ans = INF;
    for(int i = 1;i<=n;++i)
    {
        long long cnt = i,tmp;
        h_ = h + sum_[i];
        if(h_%sum==0) tmp = -h_/sum;
                 else tmp = -h_/sum + 1;
        cnt += tmp*(n);
        ans = min(ans,cnt);
    }
    printf("%lld\n",ans);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

F1

F1因为n 只有50 所以可以想到 n*n*n*n 的暴力解法即能求解

先处理出所有的前缀和差值 然后放进一个二维偏序结构体里面判断能不能塞进 然后不断地判断暴力枚举即可

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
vector<int >vt;
const int MAX_N = 55;
int arr[MAX_N];
struct node
{
    int l,r;
    node(){}
    node(int ll ,int rr)
    {
        l = ll;
        r = rr;
    }
}tmp[MAX_N*MAX_N];
bool cmp(node a,node b)
{
    if(a.r==b.r) return a.l<b.l;
    return a.r<b.r;
}
vector<pll> anss,ans_;
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,cnt = 0,ans = 0;
    scanf("%d",&n);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    for(int i = 1;i<=n;++i) arr[i]+=arr[i-1];
    for(int i = 1;i<=n;++i)
    {
        for(int j = i;j<=n;++j)
        {
            vt.push_back(arr[j]-arr[i-1]);
        }
    }
    int sz = vt.size();
    for(int i = 0;i<sz;++i)
    {
        cnt = 0;
        int now = vt[i];
        ans_.clear();
        for(int j = 1;j<=n;++j)
        {
            for(int k = j;k<=n;++k)
            {
                if(arr[k]-arr[j-1]==now)
                    tmp[++cnt]=node(j,k);
            }
        }
        sort(tmp+1,tmp+1+cnt,cmp);
        int pre = 0;
        for(int j= 1;j<=cnt;++j)
        {
            if(tmp[j].l>pre)
            {
                ans_.push_back(pll(tmp[j].l,tmp[j].r));
                pre = tmp[j].r;
            }
        }
        if(ans<ans_.size())
        {
            ans = ans_.size();
            anss = ans_;
        }
    }
    printf("%d\n",ans);
    for(int i = 0;i<ans;++i)
    printf("%d %d\n",anss[i].first,anss[i].second);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

F2

n是1500 n^3 不能做

跟第一题一样的做法 主要是你要利用第一个循环 利用 r 1 - n l 从 1 到 r 这样就能保证二维偏序 就不用sort

然后用map存出现的值 值一样的都放进一个vector

然后和第一题一样跑即可求值

#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 1525;
int arr[MAX_N];
map<int ,int > mp;
set<int > st;
vector<pll> anss,ans_,vt[MAX_N*MAX_N];
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,ans=0,cnt,CNT= 0;
    scanf("%d",&n);
    for(int i = 1;i<=n;++i) scanf("%d",&arr[i]);
    for(int i = 1;i<=n;++i) arr[i] += arr[i-1];
    for(int j = 1;j<=n;++j)
    {
        for(int i = 1;i<=j;++i)
        {
            int now = arr[j]-arr[i-1];
            if(mp.find(now)!=mp.end())  vt[mp[now]].push_back(pll(i,j));
            else mp[now] = ++CNT,vt[mp[now]].push_back(pll(i,j));
        }
    }
    map<int,int >::iterator it;
    for(it = mp.begin();it!=mp.end();it++)
    {
        int pre = 0,now = it->second;
        ans_.clear();
        cnt = 0;
        int sz = vt[now].size();
        for(int i = 0;i<sz;++i)
        {
            if(vt[now][i].first>pre)
            {
                cnt++;
                pre = vt[now][i].second;
                ans_.push_back(pll(vt[now][i].first,vt[now][i].second));
            }
        }
        if(ans<cnt)
        {
            ans = cnt;
            anss = ans_;
        }
    }
    printf("%d\n",ans);
    for(int i = 0;i<ans;++i)
        printf("%d %d\n",anss[i].first,anss[i].second);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

G

题意 给你一颗树 定义不满足的点就是他连接的所有边里面出现了相同的颜色 问你用颜色去染色 使得不满足的点数量不超过k

做法 我们知道既然你最多只能k个不满足点 那么我们贪心按照度数从大到小排序 另颜色数等于 k+1点的度数 就满足后面所有点都合法 前面所有点都是不满足点 贪心的去染色即可解决这道题

/*
        If you can't see the repay
        Why not just working step by step?
        to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
#include <random>
#include <ctime>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
//inv[1]=1;
//for(int i=2; i<M; ++i) inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
//for(int i=0; i<LIM; ++i) coef[i]=1ll*(P[i]-1)*inv[P[i]]%mod;
const int MAX_N = 200025;
int col[MAX_N],ans[MAX_N],arr[MAX_N],m;
vector<int > G[MAX_N],id[MAX_N];
struct node
{
    int u,fa;
    node(){}
    node(int uu,int faa){
        u = uu;
        fa = faa;
    }
};
bool cmp(int a,int b)
{
    return a > b;
}
void bfs()
{
    queue<node> q;
    q.push(node(1,0));
    while(!q.empty())
    {
        node now = q.front();
        q.pop();
        int cnt = col[now.u];
        int sz = G[now.u].size();
        for(int i = 0;i<sz;++i)
        {
            int to = G[now.u][i];
            if(to==now.fa) continue;
            cnt++;
            if(cnt>m) cnt-=m;
            col[to] = cnt;
            ans[id[now.u][i]] = cnt;
            q.push(node(to,now.u));
        }
    }
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,k,a,b;
    scanf("%d%d",&n,&k);
    for(int i = 1;i<n;++i)
    {
        scanf("%d%d",&a,&b);
        G[a].push_back(b);
        id[a].push_back(i);
        arr[a]++;
        G[b].push_back(a);
        id[b].push_back(i);
        arr[b]++;
    }
    sort(arr+1,arr+1+n,cmp);
    m = arr[k+1];

    bfs();
    printf("%d\n",m);
    for(int i = 1;i<n;++i)
        i==n?printf("%d\n",ans[i]):printf("%d ",ans[i]);
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}

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