Codeforces Round #494 (Div. 3)

Codeforces Round #494 (Div. 3)

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A. Polycarp's Pockets

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)

using namespace std;
int a[111],ans,n;
int main() {
    scanf("%d",&n);
    rep(i,1,n){int x;
        scanf("%d",&x);
        ++a[x];
        ans=max(ans,a[x]);
    }
    printf("%d\n",ans);
    return 0;
}

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B. Binary String Constructing

先用01交替，相邻不同的位置个数为x-1，然后把剩余的01放上去使得答案为x。第一个数可能是1，也可能是0，都要计算

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)

using namespace std;
int num[2],x,A[555],cc;
int ck(){int ans=0;
    rep(i,1,cc-1)if(A[i]!=A[i+1])++ans;
    return ans==x;
}
int main() {
    scanf("%d%d%d",&num[0],&num[1],&x);
    int a=num[0],b=num[1];
    A[++cc]=0;
    --num[0];
    for(int i=1;i<=x-1;++i){
        A[++cc]=A[cc-1]^1;
        --num[A[cc]];
    }
    int t=A[cc];
    while(num[t])A[++cc]=t,--num[t];
    while(num[t^1])A[++cc]=(t^1),--num[t^1];
    if(ck()){rep(i,1,cc)printf("%d",A[i]);return puts(""),0;}


    cc=0;
    num[0]=a;num[1]=b;
    A[++cc]=1;
    --num[1];
    for(int i=1;i<=x-1;++i){
        A[++cc]=A[cc-1]^1;
        --num[A[cc]];
    }
    t=A[cc];
    while(num[t])A[++cc]=t,--num[t];
    while(num[t^1])A[++cc]=(t^1),--num[t^1];
    if(ck()){rep(i,1,cc)printf("%d",A[i]);return puts(""),0;}

    return 0;
}

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C. Intense Heat

推了半天dp，发现是暴力。。。

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define LD long double
using namespace std;
int a[5555],n,k;
LD ans=0;
int main() {
    scanf("%d%d",&n,&k);
    rep(i,1,n) scanf("%d",&a[i]);
    rep(i,2,n) a[i]+=a[i-1];
    rep(i,1,n){
        rep(j,i+k-1,n){
            int sum = a[j]-a[i-1];
            ans = max(ans,(LD)sum/(j-i*1.0+1.0));
        }
    }
    printf("%.10f\n",(double)ans);
    return 0;
}

D. Coins and Queries

存一下，每种value出现了多少次，从大到小贪心即可

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define LD long double
typedef long long ll;
using namespace std;
int n,A[55],m;
int main() {
    scanf("%d%d",&n,&m);
    rep(i,1,n) {ll x;int tmp=0;
        scanf("%I64d",&x);
        while(x)x>>=1LL,++tmp;
        ++A[tmp-1];
    }

    rep(ti,1,m){ll x,ans=0;
        scanf("%I64d",&x);
        for(ll i=35;i>=0;--i){
            ll t = (1LL<<i);
            ll s = x/t;
            ll tmp = s*t;
            if(s>A[i])s=A[i];
            ans += s;
            x-=s*t;
        }
        printf("%I64d\n",x==0?ans:-1);
    }
    return 0;
}

E. Tree Constructing

先把一条直径构造出来，从左端开始早每个节点的分出来的树，通过节点的编号确定树高，k的确定分支的个数。构造完check一下是否合法，即是一棵树，且直径符合条件。然后我check直径忘+1了。。再快30s就A了。。。依旧是zz般的手速

#include <bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define LD long double
typedef long long ll;
const int N = 4e5 + 7;
using namespace std;
vector<int> G[N][2];
void add(int u,int v){
    G[u][0].push_back(v);
    G[v][1].push_back(u);
}
int n,d,k;
int cald(int p,int d){
    if(d%2==0){
        if(p<=d/2)return p-1;
        else return d-p;
    }
    else {
        int t = (d+1)/2;
        if(p<=t) return p-1;
        else return d-p;
    }
}
int cnt,sk[N],h[N];
void bfs(int s,int dep){
    queue<int> q;
    q.push(s);h[s]=0;
    while(!q.empty()){
        int u=q.front();q.pop();

        if(h[u]==dep)continue;
        while(sk[u]<k&&cnt<n){
            ++sk[u];
            ++cnt;
            ++sk[cnt];
            h[cnt]=h[u]+1;
            add(u,cnt);
            q.push(cnt);
        }
    }
}
int vis[N],dis[N];
int ck(){
    queue<int> q;
    q.push(1);vis[1]=1;dis[1]=1;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=0;i<G[u][0].size();++i){
            int v=G[u][0][i];
            if(vis[v])continue;
            vis[v]=1;
            dis[v]=dis[u]+1;
            q.push(v);
        }
        for(int i=0;i<G[u][1].size();++i){
            int v=G[u][1][i];
            if(vis[v])continue;
            vis[v]=1;
            dis[v]=dis[u]+1;
            q.push(v);
        }
    }
    rep(i,1,n)if(!vis[i])return 0;

    int idx=1;
    rep(i,2,n){
        if(dis[idx]<dis[i])idx=i;
    }
    q.push(idx);
    dis[idx]=1;memset(vis,0,sizeof(vis));
    vis[idx]=1;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=0;i<G[u][0].size();++i){
            int v=G[u][0][i];
            if(vis[v])continue;
            vis[v]=1;
            dis[v]=dis[u]+1;
            q.push(v);
        }
        for(int i=0;i<G[u][1].size();++i){
            int v=G[u][1][i];
            if(vis[v])continue;
            vis[v]=1;
            dis[v]=dis[u]+1;
            q.push(v);
        }
    }
    idx=1;
    rep(i,2,n){
        if(dis[idx]<dis[i])idx=i;
    }
    return dis[idx]==d;
}
int main() {
    scanf("%d%d%d",&n,&d,&k);
    d++;
    cnt = d;
    for(int i=2;i<=d;++i){
        add(i-1,i);
        if(i<d){
            int dep = cald(i,d);
            sk[i]=2;
            bfs(i,dep);
        }

    }
    if(ck()){
        puts("YES");
        rep(i,1,n){
            for(int j=0;j<G[i][0].size();++j)printf("%d %d\n",i,G[i][0][j]);
        }
    }
    else puts("NO");
    return 0;
}