HDU 全国多校第四场 题解

题解

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A AND Minimum Spanning Tree

参考代码：

#include<bits/stdc++.h>
#define maxl 200010
using namespace std;

int n,ans1;
int mi[];
int ans[maxl];

inline void prework()
{
    scanf("%d",&n);
}

inline int find(int x)
{
    for(int j=;j<=;j++)
    if((x&mi[j])==)
        return mi[j];
}

inline void mainwork()
{
    ans1=;int x;
    for(int i=;i<=n;i++)
    {
        if(i&)
        {
            x=find(i);
            if(x<=n)
                ans[i]=x;
            else
                ans1++,ans[i]=;
        }
        else
            ans[i]=;
    }
}

inline void print()
{
    printf("%d\n",ans1);
    for(int i=;i<=n;i++)
        printf("%d%c",ans[i],(i==n)?'\n':' ');
}

int main()
{
    mi[]=;
    for(int i=;i<=;i++)
        mi[i]=mi[i-]*;
    int t;
    scanf("%d",&t);
    for(int i=;i<=t;i++)
    {
        prework();
        mainwork();
        print();
    }
    return ;
}

B Colored Tree

unsolved.

C Divide the Stones

题解：https://blog.csdn.net/liufengwei1/article/details/97970041

 #include<bits/stdc++.h>
 #define maxl 100010
 using namespace std;

 long long k,n,sum,t;
 long long dy[maxl];
 long long last[maxl],to[maxl];
 long long tmp[maxl];
 bool flag;
 vector <long long> ans[maxl];

 inline void prework()
 {
     scanf("%lld%lld",&n,&k);
     sum=1ll*n*(n+)/;
     for(long long i=;i<=k;i++)
         ans[i].clear();
 }

 inline void mainwork()
 {
     flag=false;
     if(sum%k!=)
         return;
     sum=sum/k;
     long long id;
     t=n/k;
     if(t%==)
     {
         long long id=;
         for(long long i=;i<=k;i++)
         {
             for(long long j=;j<=t/;j++)
             {
                 ans[i].push_back(id);
                 ans[i].push_back(n-id+);
                 id++;
             }
         }
         flag=true;
     }
     else
     {
         if(n/k==)
         {
             if(n==)
             {
                 ans[].push_back();
                 flag=true;
             }
             return;
         }
         for(long long i=;i<=k/+;i++)
         {
             dy[i]=k/+i;
             to[i]=(i-)*+;
         }
         for(long long i=k/+;i<=k;i++)
         {
             dy[i]=i-(k/)-;
             to[i]=(i-(k/)-)*;
         }
         for(long long i=;i<=k;i++)
             ans[i].push_back(i),last[i]=i,tmp[i]=i;
         long long num;
         for(long long i=;i<n/k;i++)
         {
             for(long long j=;j<=k;j++)
             {
                 num=dy[last[j]]+(i-)*k;
                 ans[j].push_back(num);
                 last[j]=to[last[j]];
                 tmp[j]+=num;
             }
         }
         for(long long i=;i<=k;i++)
             ans[i].push_back(sum-tmp[i]);
         flag=true;
     }
 }

 inline void print()
 {
     if(flag)
     {
         puts("yes");
         long long l;
         for(long long i=;i<=k;i++)
         {
             for(long long j=;j<n/k;j++)
                 printf("%lld%c",ans[i][j],(j==(n/k-))?'\n':' ');
         }
     }
     else
         puts("no");
 }

 int main()
 {
     long long t;
     scanf("%lld",&t);
     for(long long i=;i<=t;i++)
     {
         prework();
         mainwork();
         print();
     }
     return ;
 }

D Enveloping Convex

unsolved.

E Good Numbers

unsolved.

F Horse

unsolved.

G Just an Old Puzzle

题解：百度“15难题”

参考代码

#include<bits/stdc++.h>
using namespace std;

int ans;
int a[];
int x[];

inline void prework()
{
    for(int i=;i<=;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]==)
        {
            a[i]=;
            ans=x[i];
        }
    }
}

inline void mainwork()
{
    for(int i=;i<=;i++)
    {
        for(int j=i+;j<=;j++)
        if(a[j]<a[i])
            ans++;
    }

}

inline void print()
{
    if(ans&)
        puts("No");
    else
        puts("Yes");
}

int main()
{
    x[]=;x[]=;x[]=;x[]=;
    x[]=;x[]=;x[]=;x[]=;
    int t;
    scanf("%d",&t);
    for(int cas=;cas<=t;cas++)
    {
        prework();
        mainwork();
        print();
    }
    return ;
}

H K-th Closest Distance

题解：主席树+二分 https://blog.csdn.net/liufengwei1/article/details/97948584

参考代码

#include<bits/stdc++.h>
#define maxl 100010
using namespace std;

const int nn=1e6;

int n,m,tot,ans;
int rt[maxl],a[maxl];
struct node
{
    int ls,rs,sum;
}tree[maxl**];

inline void insert(int num,int &x,int l,int r)
{
    tree[++tot]=tree[x];x=tot;
    ++tree[x].sum;
    if(l==r) return;
    int mid=(l+r)>>;
    if(num<=mid)
        insert(num,tree[x].ls,l,mid);
    else
        insert(num,tree[x].rs,mid+,r);
}

inline void prework()
{
    scanf("%d%d",&n,&m);
    for(int i=;i<=n;i++)
        scanf("%d",&a[i]);
    tree[].ls=tree[].rs=tree[].sum=;
    rt[]=;
    tot=;
    for(int i=;i<=n;i++)
    {
        rt[i]=rt[i-];
        insert(a[i],rt[i],,nn);
    }
}

inline int query(int i,int j,int l,int r,int i1,int j1)
{
    if(i1==l && j1==r)
        return tree[j].sum-tree[i].sum;
    int mid=(i1+j1)>>,ret;
    if(r<=mid)
        ret=query(tree[i].ls,tree[j].ls,l,r,i1,mid);
    else if(l>mid)
        ret=query(tree[i].rs,tree[j].rs,l,r,mid+,j1);
    else
    {
        ret=query(tree[i].ls,tree[j].ls,l,mid,i1,mid);
        ret+=query(tree[i].rs,tree[j].rs,mid+,r,mid+,j1);
    }
    return ret;
}

inline bool jug(int l,int r,int mid,int p,int k)
{
    int up=min(p+mid,nn),lo=max(,p-mid);
    int sum=query(rt[l-],rt[r],lo,up,,nn);
    if(sum<k)
        return false;
    else
        return true;
}

inline void mainwork()
{
    ans=;int up,lo,p,k,l,r,mid;
    for(int i=;i<=m;i++)
    {
        scanf("%d%d%d%d",&lo,&up,&p,&k);
        lo^=ans;up^=ans;p^=ans;k^=ans;
        l=;r=nn;
        while(l+<r)
        {
            mid=(l+r)>>;
            if(!jug(lo,up,mid,p,k))// <k
                l=mid;
            else
                r=mid;
        }
        if(jug(lo,up,l,p,k))
            ans=l;
        else
            ans=l+;
        printf("%d\n",ans);
    }
}

inline void print(){}

int main()
{
    int t;
    scanf("%d",&t);
    for(int i=;i<=t;i++)
    {
        prework();
        mainwork();
        print();
    }
    return ;
}

I Linear Functions

unsolved.

J Minimal Power of Prime

参考代码

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int size=1e6+;
double eps=1e-;
int p[size];bool prime[size];
int mpri[size];
int tot=;
void init()
{
    for(int i=;i<size;i++) prime[i]=true;
    for(int i=;i<size;i++)
    {
        if(prime[i])
        {
            p[++tot]=i;
            mpri[i]=i;
        }
        for(int j=;j<=tot&&p[j]*i<size;j++)
        {
            prime[i*p[j]]=false;mpri[i*p[j]]=p[j];
            if(i%p[j]==) break;
        }
    }
}
int main()
{
    init();
    int t;
    long long x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&x);
        int cnt=;
        int ans=;
        if(x<size){
            int ps=mpri[x];
            while(x!=)
            {
                do x/=ps,cnt++;
                while(mpri[x]==ps);
                ps=mpri[x];
                ans=min(ans,cnt);
            }
            printf("%d\n",ans);
            continue;
        }
        bool flag=false;
        for(int i=;i<=tot;i++)
        {
            cnt=;
            if(x%p[i]==)
            {
                do x/=p[i],cnt++;
                while(x%p[i]==);
            }
            if(cnt==)
            {
                puts("");
                flag=true;
                break;
            }
        }
        if(flag) continue;
        LL sq=sqrt(x)+eps;
        if(sq*sq==x)
        {
            printf("2\n");
        }
        else
        {
            puts("");
        }
    }
}    

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