Codeforces Round #615 (Div. 3)补题

题目

A - Collecting Coins

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
    ll t;
    scanf("%lld",&t);
    while(t--){
        ll a,b,c,n;
        scanf("%lld%lld%lld%lld",&a,&b,&c,&n);
        ll MAX=max(a,max(b,c));
        ll x=MAX*3-a-b-c;
        if(n<x||(n-x)%3)puts("NO");
        else puts("YES");
    }
}

B - Collecting Packages

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<pair<int,int> >vec;
int main(){
    ll t;
    scanf("%lld",&t);
    ll n;
    while(t--){
        scanf("%lld",&n);
        vec.clear();
        for(ll i=1;i<=n;i++){
            ll x,y;
            scanf("%lld%lld",&x,&y);
            vec.push_back(make_pair(x,y));
        }
        sort(vec.begin(),vec.end());
        bool flag=true;
        for(ll i=0;i<vec.size()-1;i++){
            if(vec[i].second>vec[i+1].second){
                flag=false;
                break;
            }
        }
        if(!flag){
            puts("NO");
            continue;
        }
        else{
            puts("YES");
            ll curx=0,cury=0;
            for(ll i=0;i<vec.size();i++){
                while(curx<vec[i].first){
                    curx++;
                    printf("R");
                }
                while(cury<vec[i].second){
                    cury++;
                    printf("U");
                }
            }
            printf("\n");
        }
    }
}

C - Product of Three Numbers

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
bool notprime[maxn];
int main(){
    ll t;
    scanf("%lld",&t);
    for(ll i=2;i<maxn;i++){
        if(!notprime[i]){
            ll x=i*i;
            while(x<maxn){
                notprime[x]=true;;
                x*=x;
            }
        }
    }
    while(t--){
        vector<ll>vec;
        ll n;
        scanf("%lld",&n);
        for(ll i=2;i*i<=n;i++){
            if(!notprime[i]){
                ll x=n;
                ll cur=0;
                while(x%i==0){
                    cur++;
                    if(cur==1){
                        vec.push_back(i);
                    }
                    else if(cur==3){
                        vec.push_back(i*i);
                    }
                    else if(cur==6){
                        vec.push_back(i*i*i);
                    }
                    x/=i;
                }
                if(vec.size()>=3)break;
            }
        }
        if(vec.size()>=3){
            puts("YES");
            cout<<vec[0]<<' '<<vec[1]<<' '<<n/vec[0]/vec[1]<<endl;
        }
        else if(vec.size()==2){
                ll x=n/vec[1]/vec[0];
            if(x!=vec[0]&&x!=vec[1]&&x!=1){
            puts("YES");
            cout<<vec[0]<<vec[1]<<x<<endl;
            }
            else{
                puts("NO");
            }
        }
        else puts("NO");
    }
}

D - MEX maximizing

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=0x3f3f3f3f;
map<ll,ll>mp;
set<pair<ll,ll> >se;
 
int main(){
    ll q,x;
    scanf("%lld%lld",&q,&x);
    for(ll i=0;i<x;i++){
        se.insert(make_pair(0,i));
    }
    for(ll i=1;i<=q;i++){
        ll a;
        scanf("%lld",&a);
        se.erase(make_pair(mp[a%x],a%x));
        mp[a%x]++;
        se.insert(make_pair(mp[a%x],a%x));
    printf("%lld\n",se.begin()->first*x+se.begin()->second);
    }
}

E - Obtain a Permutation

对于每一个数判断是否可以翻转到应该在的位置，若可以，则这一列翻转对应次数的代价减一，然后求每一列的最小代价，相加为答案。

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
const int INF=0x3f3f3f3f;
vector<int>vec[maxn];
map<int,int>cnt[maxn];
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++){
            int x;
            scanf("%d",&x);
            vec[i].push_back(x);
            if((x-1)%m==j&&x<=m*n)
                cnt[j][(i+n-(x-1)/m)%n]++;
        }
    }
    int ans=0;
    for(int j=0;j<m;j++){
    int cur=INF;
    for(int i=0;i<n;i++){
        cur=min(cur,n-cnt[j][i]+i);
    }
    ans+=cur;
    }
    cout<<ans<<endl;
}

F - Three Paths on a Tree

两次dfs找树的直径，再对直径所有点多起点bfs找最远点

#include<bits/stdc++.h>
using namespace std;
const int maxn=2e5+10;
int head[maxn];
int fa[maxn];
int d[maxn];
bool vis[maxn];
struct Edge{
    int v,next;
}e[maxn<<1];
int cnt;
void add(int u,int v){
    e[cnt].v=v;
    e[cnt].next=head[u];
    head[u]=cnt++;
}
int p1,p2,p3;
int P;
int maxd;
void dfs(int u,int f){
    fa[u]=f;
    d[u]=d[f]+1;
    if(d[u]>maxd)P=u,maxd=d[u];
    for(int i=head[u];~i;i=e[i].next){
        int v=e[i].v;
        if(v==f)continue;
        dfs(v,u);
    }
}
queue<pair<int,int> >q;
int main(){
    memset(head,-1,sizeof(head));
    int n;
    scanf("%d",&n);
    for(int i=1;i<n;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v);
        add(v,u);
    }
    maxd=0;
    dfs(1,0);
    p1=P;
    maxd=0;
    dfs(p1,0);
    p2=P;
    q.push(make_pair(p1,0));
    vis[p1]=1;
    int r=0;
    while(P!=p1){
        vis[P]=1;
        q.push(make_pair(P,0));
        P=fa[P];
        r++;
    }
    int dis=0;
    while(!q.empty()){
        pair<int,int>x=q.front();
        q.pop();
        p3=x.first;
        dis=x.second;
        for(int i=head[p3];~i;i=e[i].next){
            int v=e[i].v;
            if(!vis[v]){
                vis[v]=1;
                q.push(make_pair(v,x.second+1));
            }
        }
    }
    printf("%d\n",dis+r);
    printf("%d %d %d\n",p1,p2,p3);
}