Educational Codeforces Round 58 Div. 2 自闭记

时间：2023-03-09 19:20:02
Educational Codeforces Round 58 Div. 2 自闭记

　　明明多个几秒就能场上AK了。自闭。

　　A：签到。

#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 T,l,r,d;

signed main()

{

/*#ifndef ONLINE_JUDGE

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

    const char LL[]="%I64d\n";

#endif*/

    T=read();

    while (T--)

    {

        l=read(),r=read(),d=read();

        if (d<l) cout<<d<<endl;

        else cout<<1ll*(r/d+)*d<<endl;

    }

    return ;

    //NOTICE LONG LONG!!!!!

}

　　B：签到。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

#define ll long long

#define N 500010

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;

char s[N];

signed main()

{

/*#ifndef ONLINE_JUDGE

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

    const char LL[]="%I64d\n";

#endif*/

    scanf("%s",s+);n=strlen(s+);

    bool flag=;int x=n+,y=;

    for (int i=;i<=n;i++)

    {

        if (s[i]=='[') flag=;

        if (s[i]==':') if (flag) {x=i;break;}

    }

    flag=;

    for (int i=n;i>=;i--)

    {

        if (s[i]==']') flag=;

        if (s[i]==':') if (flag) {y=i;break;}

    }

    if (x>=y) cout<<-;

    else

    {

        int ans=;

        for (int i=x+;i<y;i++)

        if (s[i]=='|') ans++;

        cout<<ans;

    }

    return ;

    //NOTICE LONG LONG!!!!!

}

　　C：wa了无数发。按左端点排序后找一个连续且和其他线段不相交的线段集即可。

#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 T,n,ans[N];

struct data

{

    int l,r,i;

    bool operator <(const data&a) const

    {

        return l<a.l;

    }

}a[N];

signed main()

{

#ifndef ONLINE_JUDGE

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

    const char LL[]="%I64d\n";

#endif

    T=read();

    while (T--)

    {

        n=read();

        for (int i=;i<=n;i++) a[i].l=read(),a[i].r=read(),a[i].i=i;

        sort(a+,a+n+);

        bool flag=;a[n+].l=;

        int t=,x=a[].r;

        while (t<n&&a[t+].l<=x) t++,x=max(x,a[t].r);

        if (t==n) cout<<-<<endl;

        else

        {

            for (int i=;i<=t;i++) ans[a[i].i]=;

            for (int i=t+;i<=n;i++) ans[a[i].i]=;

            for (int i=;i<=n;i++) printf("%d ",ans[i]);

            cout<<endl;

        }

    }

    return ;

    //NOTICE LONG LONG!!!!!

}

　　E：这才是真签到吧？wa了一发自闭啊？

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

using namespace std;

#define ll long long

#define N 500010

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 m,u,v;

signed main()

{

#ifndef ONLINE_JUDGE

    freopen("b.in","r",stdin);

    freopen("b.out","w",stdout);

    const char LL[]="%I64d\n";

#endif

    m=read();

    while (m--)

    {

        char c=getchar();while (c!='+'&&c!='?') c=getchar();

        int x=read(),y=read();if (x>y) swap(x,y);

        if (c=='+')

        {

            u=max(u,x),v=max(y,v);

        }

        else

        {

            if (x>=u&&y>=v) printf("YES\n");

            else printf("NO\n");

        }

    }

    return ;

    //NOTICE LONG LONG!!!!!

}

　　D：如果路径经过某点，最后所得的路径gcd显然是该点某些质因子的倍数。对此dp即可。一发wa on 3，3可是样例啊？自闭了啊？

#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,a[N],p[N],f[N][],prime[N][],t,ans;

struct data{int to,nxt;

}edge[N<<];

void addedge(int x,int y){t++;edge[t].to=y,edge[t].nxt=p[x],p[x]=t;}

void dfs(int k,int from)

{

    for (int i=p[k];i;i=edge[i].nxt)

    if (edge[i].to!=from)

    {

        dfs(edge[i].to,k);

        for (int x=;x<=prime[edge[i].to][];x++)

            for (int y=;y<=prime[k][];y++)

            if (prime[edge[i].to][x]==prime[k][y])

            {

                ans=max(ans,f[k][y]+f[edge[i].to][x]+);

                f[k][y]=max(f[k][y],f[edge[i].to][x]+);

            }

    }

}

signed main()

{

#ifndef ONLINE_JUDGE

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

    const char LL[]="%I64d\n";

#endif

    n=read();

    for (int i=;i<=n;i++) a[i]=read();

    bool flag=;

    for (int i=;i<=n;i++) if (a[i]!=) flag=;

    if (flag) {cout<<;return ;}

    for (int i=;i<n;i++)

    {

        int x=read(),y=read();

        addedge(x,y),addedge(y,x);

    }

    for (int i=;i<=n;i++)

    {

        for (int j=;j*j<=a[i];j++)

        if (a[i]%j==)

        {

            prime[i][++prime[i][]]=j;

            while (a[i]%j==) a[i]/=j;

        }

        if (a[i]>) prime[i][++prime[i][]]=a[i];

    }

    dfs(,);

    cout<<ans+;

    return ;

    //NOTICE LONG LONG!!!!!

}

　　G：一眼线性基，然后就往别的方面想了。自闭了半天直接乱搞求个前缀异或和搞了个线性基上去就pp了。冷静了半天正确性何在。事实上划分序列相当于选出一些前缀异或和，这是一个裸到不行的线性基。注意虽然最后一个前缀和应该是必须选的，但是不考虑也不会造成什么影响（吧）。

#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,a[N],base[],ans;

signed main()

{

#ifndef ONLINE_JUDGE

    freopen("a.in","r",stdin);

    freopen("a.out","w",stdout);

    const char LL[]="%I64d\n";

#endif

    n=read();

    for (int i=;i<=n;i++) a[i]=a[i-]^read();

    if (a[n]==) {cout<<-;return ;}

    for (int i=;i<=n;i++)

        for (int j=;~j;j--)

        if (a[i]&(<<j))

        {

            if (base[j]) a[i]^=base[j];

            else {base[j]=a[i],ans++;break;}

        }

    cout<<ans;

    return ;

    //NOTICE LONG LONG!!!!!

}

　　F：随便都知道是二分答案。但是nmlog显然有些吃力。考虑random_shuffle一发，每次记录当前需要的最大容量，考虑下一辆卡车时先判断当前容量是否能满足其需求，如果不行再二分一下。这样复杂度大约是nm+nlogmlogV，因为只需要对所需容量的单调栈中的卡车进行二分，而排列又是随机的。复杂度证明似乎在一篇cfblog里看到过。（突然发现整个idea都在这篇blog里https://codeforces.com/blog/entry/62602）一开始又没注意到要开long long，交一发in queue了半天，然后wa on 3，结果改了一发又没改全，交一发再次in queue了半天，接着wa on 3。然后就只剩20s了，手速不行，就，自闭了。

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstdlib>

#include<cstring>

#include<algorithm>

#include<ctime>

using namespace std;

#define ll long long

#define N 410

#define M 250010

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,a[N];

ll ans;

struct data{int s,f,c,r;

}b[M];

bool check(ll k,int i)

{

    ll cur=k;int cnt=;

    for (int j=b[i].s;j<b[i].f;j++)

    {

        if (cur>=1ll*b[i].c*(a[j+]-a[j])) cur-=1ll*b[i].c*(a[j+]-a[j]);

        else

        {

            cur=k,cnt++;

            if (cur>=1ll*b[i].c*(a[j+]-a[j])) cur-=1ll*b[i].c*(a[j+]-a[j]);

            else return ;

        }

        if (cnt>b[i].r) return ;

    }

    return ;

}

signed main()

{

#ifndef ONLINE_JUDGE

    freopen("b.in","r",stdin);

    freopen("b.out","w",stdout);

    const char LL[]="%I64d\n";

#endif

    srand(time());

    n=read(),m=read();

    for (int i=;i<=n;i++) a[i]=read();

    for (int i=;i<=m;i++) b[i].s=read(),b[i].f=read(),b[i].c=read(),b[i].r=read();

    random_shuffle(b+,b+m+);

    for (int i=;i<=m;i++)

    if (!check(ans,i))

    {

        ll l=ans+,r=1000000000000000000ll;

        while (l<=r)

        {

            ll mid=l+r>>;

            if (check(mid,i)) ans=mid,r=mid-;

            else l=mid+;

        }

    }

    cout<<ans;

    return ;

    //NOTICE LONG LONG!!!!!

}

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