Dwango Programming Contest V 翻车记

　　A：签到。

#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<'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],s,ans=0;
int calc(int k){return abs(s-a[k]*n);}
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif*/
    n=read();
    for (int i=1;i<=n;i++) s+=a[i]=read();
    for (int i=2;i<=n;i++)
    if (calc(i)<calc(ans+1)) ans=i-1;
    cout<<ans;
    return 0;
}

　　B：按位从高到低贪心，对当前位数一下有多少个子区间和该位为1，如果能满足要求就将其统计入答案并删掉所有该位为0的子区间和。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1010
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,b[N],cnt=0;
ll a[N*N],c[N*N],ans;
int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("a.in","r",stdin);
    freopen("a.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif*/
    n=read(),m=read();
    for (int i=1;i<=n;i++) b[i]=read();
    for (int i=1;i<=n;i++)
    {
        a[++cnt]=b[i];
        for (int j=i+1;j<=n;j++)
        cnt++,a[cnt]=a[cnt-1]+b[j];
    }
    n=cnt;
    for (int j=40;~j;j--)
    {
        int s=0;
        for (int i=1;i<=n;i++)
        if (a[i]&(1ll<<j)) c[++s]=a[i];
        if (s>=m)
        {
            ans+=1ll<<j;n=s;
            for (int i=1;i<=s;i++) a[i]=c[i];
        }
    }
    cout<<ans;
    return 0;
}

　　C：指针从左到右扫一遍，两个队列分别维护指针左方D的出现位置和右方C的出现位置，并记录跨过当前位置的合法D~C区间有多少个。如果指针扫到M就更新答案，扫到D将其加入左边队列并统计以该位置为左端点的合法D~C区间数量，扫到C将其弹出右边队列并统计以该位置为右端点的合法D~C区间数量，拿两个指针记录两个队列中能产生贡献的边界位置即可做到线性。为啥atcoder不define ONLINE_JUDGE啊白交了一发。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 1000010
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,q,m,a[N],b[N],c[N];
char s[N];
int main()
{
    n=read();scanf("%s",s+1);
    for (int i=1;i<=n;i++)
    if (s[i]=='D') a[i]=0;
    else if (s[i]=='M') a[i]=1;
    else if (s[i]=='C') a[i]=2;
    else a[i]=-1;
    q=read();
    while (q--)
    {
        m=read();
        int hb=1,tb=0,hc=1,t=0,tc=0;ll ans=0,cur=0;
        for (int i=1;i<=n;i++) if (a[i]==2) b[++tb]=i;
        for (int i=1;i<=n;i++)
        if (a[i]==1) ans+=cur;
        else if (a[i]==0)
        {
            c[++tc]=i;
            while (t<tb&&b[t+1]-i+1<=m) t++;
            cur+=t-hb+1; 
        }
        else if (a[i]==2)
        {
            while (hc<=tc&&i-c[hc]+1>m) hc++;
            cur-=tc-hc+1;
            hb++;
        }
        cout<<ans<<endl;
    }
    return 0;
}

　　D：可以发现每个点最终可以被移动到该剩余类的任意位置。那么数一下每个剩余类有多少个点。这些点最后应该排成a*(a-1)的长方形或a*a的正方形。可以枚举最终矩形左下角坐标并二分边长check一发，这样就能拿到部分分了。然后好像还不太会。

　　E：没看。

　　感觉atcoder赛制不是非常友好啊。result：rank 185 rating +352