Avito Cool Challenge 2018 自闭记

　　A：n==2?2:1。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    if (n==) cout<<;else cout<<;
    return ;
}

　　B：sort一下，组内人数相同的放一组，超了就换一组。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,ans[N],cnt;
struct data
{
    int x,y;
    bool operator <(const data&a) const
    {
        return x<a.x;
    }
}a[N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    for (int i=;i<=n;i++) a[i].x=n-read(),a[i].y=i;
    sort(a+,a+n+);
    for (int i=;i<=n;i++)
    {
        int t=i;
        while (t<n&&a[t+].x==a[i].x&&t+-i+<=a[i].x) t++;
        if (t-i+!=a[i].x) {cout<<"Impossible";return ;}
        cnt++;
        for (int j=i;j<=t;j++) ans[a[j].y]=cnt;
        i=t;
    }
    printf("Possible\n");
    for (int i=;i<=n;i++) printf("%d ",ans[i]);
    return ;
}

　　C：f[i][j]前i个有j块那啥的方案数，转移显然。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 998244353
#define N 2010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,k,f[N][N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read(),m=read(),k=read();
    f[][]=m;
    for (int i=;i<=n;i++)
    {
        f[i][]=m;
        for (int j=;j<=k;j++)
        f[i][j]=(f[i-][j]+1ll*f[i-][j-]*(m-))%P;
    }
    cout<<f[n][k];
    return ;
}

　　D：kruskal求出最小瓶颈树，最后使所有标记点连接成一个连通块的边即为答案。莫名其妙的写了一堆东西还wa了两发。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,m,k,fa[N];
bool flag[N],f[N];
struct data
{
    int x,y,z;
    bool operator <(const data&a) const
    {
        return z<a.z;
    }
}e[N];
namespace tree
{
    int p[N],t,size[N],dfn[N],cnt;
    struct data{int to,nxt,len;
    }edge[N<<];
    void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
    void dfs(int k,int from)
    {
        size[k]=f[k];dfn[k]=++cnt;
        for (int i=p[k];i;i=edge[i].nxt)
        if (edge[i].to!=from)
        {
            dfs(edge[i].to,k);
            size[k]+=size[edge[i].to];
        }
    }
}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read(),m=read(),k=read();
    for (int i=;i<=k;i++) f[read()]=;
    /*for (int i=1;i<=m;i++)
    {
        int x=read(),y=read(),z=read();
        addedge(x,y,z),addedge(y,x,z);
    }*/
    for (int i=;i<=m;i++) e[i].x=read(),e[i].y=read(),e[i].z=read();
    sort(e+,e+m+);
    for (int i=;i<=n;i++) fa[i]=i;
    for (int i=;i<=m;i++)
    {
        int p=find(e[i].x),q=find(e[i].y);
        if (p!=q) fa[q]=p,tree::addedge(e[i].x,e[i].y,e[i].z),tree::addedge(e[i].y,e[i].x,e[i].z),flag[i]=;
    }
    tree::dfs(,);
    for (int i=m;i>=;i--)
    if (flag[i])
    {
        int x=e[i].x,y=e[i].y;
        if (tree::dfn[x]<tree::dfn[y]) swap(x,y);
        if (tree::size[x]&&tree::size[x]<k) {for (int j=;j<=k;j++) printf("%d ",e[i].z);return ;}
    }
    return ;
}

　　E：真的自闭了。本身就非常思博还想了50min。肝F无果回来hack的时候，在最后1min发现了输出没开I64d，居然还犹豫了一下是不是实际上并不会出问题，然后还真的就没改。看来就算define int long long也解决不了我天天爆int了。

　　注意到对于偶数位我们应该让这里的前缀和尽量小，因为下一位可以任意取。于是记录到每个偶数位时的最小前缀和即可。同时由平方差公式，可以暴力枚举偶数位上的数的因数来快速枚举每种情况。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define N 200010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,f[N];
const double eps=1E-;
ll a[N];
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
#endif*/
    n=read();
    for (int i=;i<=n/;i++) a[i<<]=read();
    for (int i=;i<=n;i+=)
    {
        for (int j=;j*j<a[i];j++)
        if (a[i]%j==)
        {
            int x=j,y=a[i]/j;//q-p=x q+p=y
            if ((x+y&)||(y-x&)) continue;
            int q=x+y>>,p=y-x>>;//p i-1ǰ׺ q iǰ׺
            if (f[i-]<p) f[i]=q,a[i-]=1ll*p*p-1ll*f[i-]*f[i-];
        }
        if (f[i]==) {cout<<"No";return ;}
    }
    cout<<"Yes\n";
    for (int i=;i<=n;i++) printf("%I64d ",a[i]);
    return ;
}

　　一夜回到解放前。虽然看起来大号还是不会变成小号的。

　　不过好像D和E都fst了不少？

　　然后我还发现F原来是真随机本来一直在想靠谱做法。

　　再也不打Chinese Round了

　　result：rank 477 rating -52