Codeforces Round #521 Div. 3 玩耍记

　　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;
int main()
{
    T=read();
    while (T--)
    {
        int a=read(),b=read(),k=read();
        if (k&) cout<<1ll*(a-b)*(k>>)+a<<endl;
        else cout<<1ll*(a-b)*(k>>)<<endl;
    }
    return ;
}

　　B：将能提供2贡献的位置先取反再扫一遍即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 110
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],ans=;
int main()
{
    n=read();
    for (int i=;i<=n;i++) a[i]=read();
    for (int i=;i<=n-;i++) if (a[i]==&&a[i-]==&&a[i+]==&&a[i-]==&&a[i+]==) ans++,a[i]=;
    for (int i=;i<n;i++) if (a[i]==&&a[i-]==&&a[i+]==) ans++,a[i+]=;
    cout<<ans;
    return ;
}

　　C：桶记录出现数字次数，枚举删掉的数字即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 200010
#define M 1000010
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],b[N],tot[M],cnt;
ll sum;
int main()
{
    n=read();
    for (int i=;i<=n;i++) sum+=a[i]=read(),tot[a[i]]++;
    for (int i=;i<=n;i++)
    {
        ll x=sum-a[i];
        if (x<=&&x%==&&tot[x/]>(a[i]==x/)) b[++cnt]=i;
    }
    cout<<cnt<<endl;
    for (int i=;i<=cnt;i++) printf("%d ",b[i]);
    return ;
}

　　D：二分答案贪心选取即可。

#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,cnt[N],id[N],ans[N];
struct data
{
    int x,y;
    bool operator <(const data&a) const
    {
        return y>a.y;
    }
}a[N],b[N];
bool check(int k)
{
    for (int i=;i<=;i++) b[i]=a[i];
    int s=;
    for (int i=;i<=;i++)
    while (b[i].y>=k&&s<m) id[++s]=b[i].x,b[i].y-=k;
    return s>=m;
}
int main()
{
    n=read(),m=read();
    for (int i=;i<=n;i++) cnt[read()]++;
    for (int i=;i<=;i++) a[i].x=i,a[i].y=cnt[i];
    sort(a+,a+);
    int l=,r=n/m;
    while (l<=r)
    {
        int mid=l+r>>;
        if (check(mid)) {for (int i=;i<=m;i++) ans[i]=id[i];l=mid+;}
        else r=mid-;
    }
    for (int i=;i<=m;i++) printf("%d ",ans[i]);
    return ;
}

　　E：枚举序列长度（显然是log级别的），二分首项大小，贪心选取即可。

#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],b[N],cnt[N],tot;
bool check(int k,int s)
{
    int t=;
    for (int i=;i<=n;i++)
    {
        if (a[i]>=k) t++,k<<=;
        if (t==s) return ;
    }
    return ;
}
int main()
{
    n=read();
    for (int i=;i<=n;i++) b[i]=a[i]=read();
    sort(b+,b+n+);
    int t=unique(b+,b+n+)-b-;
    for (int i=;i<=n;i++) a[i]=lower_bound(b+,b+t+,a[i])-b;
    for (int i=;i<=n;i++) cnt[a[i]]++;
    sort(cnt+,cnt+t+);
    for (int i=;i<=t;i++) a[i]=cnt[i];
    int tmp=n;
    n=t;
    for (int i=;i<=;i++)
    {
        int l=,r=tmp,ans=;
        while (l<=r)
        {
            int mid=l+r>>;
            if (check(mid,i)) ans=mid,l=mid+;
            else r=mid-;
        }
        tot=max(tot,ans*((<<i)-));
    }
    cout<<tot;
    return ;
}

　　F1：f[i][j]表示前i个数选取了j个且第i个被选的最大价值，转移枚举上次选哪个即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 5010
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,a[N];
ll f[N][N];
int main()
{
    n=read(),k=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    memset(f,,sizeof(f));
    f[][]=;
    for (int i=;i<=n;i++)
    {
        for (int j=;j<=m;j++)
        {
            for (int x=i-;x>=max(,i-k);x--)
            f[i][j]=max(f[i][j],f[x][j-]+a[i]);
        }
    }
    for (int i=n-;i>=max(,n-k+);i--) f[n][m]=max(f[n][m],f[i][m]);
    if (f[n][m]<) cout<<-;
    else cout<<f[n][m];
    return ;
}

　　F2：在F1基础上单调队列优化即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 5010
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,a[N],q[N],head,tail;
ll f[N][N];
int main()
{
    n=read(),k=read(),m=read();
    for (int i=;i<=n;i++) a[i]=read();
    memset(f,,sizeof(f));
    f[][]=;
    for (int j=;j<=m;j++)
    {
        head=,tail=;q[]=;
        for (int i=;i<=n;i++)
        {
            while (head<tail&&q[head]<i-k) head++;
            f[i][j]=f[q[head]][j-]+a[i];
            while (head<=tail&&f[q[tail]][j-]<=f[i][j-]) tail--;
            q[++tail]=i;
        }
    }
    for (int i=n-;i>=max(,n-k+);i--) f[n][m]=max(f[n][m],f[i][m]);
    if (f[n][m]<) cout<<-;
    else cout<<f[n][m];
    return ;
}

　　小号打的。result：rank 3 rating +311