Educational Codeforces Round 58 Div. 2 自闭记

Posted gloid

tags:

篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了Educational Codeforces Round 58 Div. 2 自闭记相关的知识，希望对你有一定的参考价值。

　　明明多个几秒就能场上AK了。自闭。

　　A：签到。

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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
";
#endif*/
    T=read();
    while (T--)
    {
        l=read(),r=read(),d=read();
        if (d<l) cout<<d<<endl;
        else cout<<1ll*(r/d+1)*d<<endl;
    }
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　B：签到。

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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
";
#endif*/
    scanf("%s",s+1);n=strlen(s+1);
    bool flag=0;int x=n+1,y=0;
    for (int i=1;i<=n;i++)
    {
        if (s[i]==‘[‘) flag=1;
        if (s[i]==‘:‘) if (flag) {x=i;break;}
    }
    flag=0;
    for (int i=n;i>=1;i--)
    {
        if (s[i]==‘]‘) flag=1;
        if (s[i]==‘:‘) if (flag) {y=i;break;}
    }
    if (x>=y) cout<<-1;
    else
    {
        int ans=4;
        for (int i=x+1;i<y;i++)
        if (s[i]==‘|‘) ans++;
        cout<<ans;
    }
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　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。

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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
";
#endif
    T=read();
    while (T--)
    {
        n=read();
        for (int i=1;i<=n;i++) a[i].l=read(),a[i].r=read(),a[i].i=i;
        sort(a+1,a+n+1);
        bool flag=0;a[n+1].l=100000000;
        int t=1,x=a[1].r;
        while (t<n&&a[t+1].l<=x) t++,x=max(x,a[t].r);
        if (t==n) cout<<-1<<endl;
        else
        {
            for (int i=1;i<=t;i++) ans[a[i].i]=1;
            for (int i=t+1;i<=n;i++) ans[a[i].i]=2;
            for (int i=1;i<=n;i++) printf("%d ",ans[i]);
            cout<<endl;
        }
    }
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　E：这才是真签到吧？wa了一发自闭啊？

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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
";
#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
");
            else printf("NO
");
        }
    }
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　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？

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,a[N],p[N],f[N][30],prime[N][30],t,ans;
struct data{int to,nxt;
}edge[N<<1];
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=1;x<=prime[edge[i].to][0];x++)
            for (int y=1;y<=prime[k][0];y++)
            if (prime[edge[i].to][x]==prime[k][y])
            {
                ans=max(ans,f[k][y]+f[edge[i].to][x]+1);
                f[k][y]=max(f[k][y],f[edge[i].to][x]+1);
            }
    }
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d
";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=read();
    bool flag=1;
    for (int i=1;i<=n;i++) if (a[i]!=1) flag=0;
    if (flag) {cout<<0;return 0;}
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        addedge(x,y),addedge(y,x);
    }
    for (int i=1;i<=n;i++)
    {
        for (int j=2;j*j<=a[i];j++)
        if (a[i]%j==0)
        {
            prime[i][++prime[i][0]]=j;
            while (a[i]%j==0) a[i]/=j;
        }
        if (a[i]>1) prime[i][++prime[i][0]]=a[i];
    }
    dfs(1,1);
    cout<<ans+1;
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　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）。

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,a[N],base[32],ans;
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d
";
#endif
    n=read();
    for (int i=1;i<=n;i++) a[i]=a[i-1]^read();
    if (a[n]==0) {cout<<-1;return 0;}
    for (int i=1;i<=n;i++)
        for (int j=30;~j;j--)
        if (a[i]&(1<<j))
        {
            if (base[j]) a[i]^=base[j];
            else {base[j]=a[i],ans++;break;}
        }
    cout<<ans;
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

　　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。

技术分享图片 
#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<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
    while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),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=0;
    for (int j=b[i].s;j<b[i].f;j++)
    {
        if (cur>=1ll*b[i].c*(a[j+1]-a[j])) cur-=1ll*b[i].c*(a[j+1]-a[j]);
        else
        {
            cur=k,cnt++;
            if (cur>=1ll*b[i].c*(a[j+1]-a[j])) cur-=1ll*b[i].c*(a[j+1]-a[j]);
            else return 0;
        }
        if (cnt>b[i].r) return 0;
    }
    return 1;
}
signed main()
{
#ifndef ONLINE_JUDGE
    freopen("b.in","r",stdin);
    freopen("b.out","w",stdout);
    const char LL[]="%I64d
";
#endif
    srand(time(0));
    n=read(),m=read();
    for (int i=1;i<=n;i++) a[i]=read();
    for (int i=1;i<=m;i++) b[i].s=read(),b[i].f=read(),b[i].c=read(),b[i].r=read();
    random_shuffle(b+1,b+m+1);
    for (int i=1;i<=m;i++)
    if (!check(ans,i))
    {
        ll l=ans+1,r=1000000000000000000ll;
        while (l<=r)
        {
            ll mid=l+r>>1;
            if (check(mid,i)) ans=mid,r=mid-1;
            else l=mid+1;
        }
    }
    cout<<ans;
    return 0;
    //NOTICE LONG LONG!!!!!
}
View Code

 

以上是关于Educational Codeforces Round 58 Div. 2 自闭记的主要内容，如果未能解决你的问题，请参考以下文章

Educational Codeforces Round 7 A

Educational Codeforces Round 7

Educational Codeforces Round 90

Educational Codeforces Round 33

Codeforces Educational Codeforces Round 54 题解

Educational Codeforces Round 27