牛客网round1

题解：

二分答案之后判断

把式子移项使得x,y不关联

#include  <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for (rint i=h;i<=t;i++)
#define dep(i,t,h) for (rint i=t;i>=h;i--)
const int N=3e5;
int a[N],sum[N],n,m;
const int INF=1e9;
bool check(int x)
{
  sum[0]=0;
  int mina=-INF;
  rep(i,1,n)
  {
    if (a[i]<x) sum[i]=sum[i-1]+1; else sum[i]=sum[i-1];
    if (i-m+1>0) mina=max(mina,2*sum[i-m]-(i-m+1));
    if (2*sum[i]-i<=mina) return(1);
  }
  return(0);
}
int main()
{
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout);
  ios::sync_with_stdio(false);
  cin>>n>>m;
  rep(i,1,n) cin>>a[i];
  int h=0,t=INF;
  while (h<t)
  {
    int mid=(h+t+1)>>1;
    if (check(mid)) h=mid; else t=mid-1;
  }
  cout<<h<<endl;
  return 0;
}

数位dp

考虑10个数字一共有18个

C(27,9) 算一下发现不大

然后就直接记忆化搜索就行了

数比较大可以用map记录

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但好像比较慢于是改了hash

然后有3个点是考细节的

后面两个是map如果初值为0就炸了可能之后一直为0 因为y可以是-1

另外一个是0要特殊考虑dfs(0)

#include <bits/stdc++.h>
using namespace std;
#define rint register int
#define IL inline
#define rep(i,h,t) for (rint i=h;i<=t;i++)
#define dep(i,t,h) for (rint i=t;i>=h;i--) 
#define ll long long
#define mp(x,y) make_pair(x,y)
ll l,n,m,f[100];
int cnt=0;
const int mo=6e6+7;
struct re{
  bool a,b;
  short c; 
  ll d,ans;
}hs[mo+1000];
queue<int> q; 
struct hash{
  ll find(bool a,bool b,short x,ll y)
  {
    int t1=((1ll*x%mo*y%mo)%mo+mo)%mo;
    while (hs[t1].c&&!(hs[t1].a==a&&hs[t1].b==b&&hs[t1].c==x&&hs[t1].d==y))
      t1++,cnt++;
    if (hs[t1].c) return(hs[t1].ans);
    return(-1);
  }
  void insert(bool a,bool b,short x,ll y,ll ans)
  {
    int t1=((1ll*x%mo*y%mo)%mo+mo)%mo;
    while (hs[t1].c) t1++,cnt++;
    hs[t1].a=a; hs[t1].b=b; hs[t1].c=x; hs[t1].d=y; hs[t1].ans=ans;
    q.push(t1);
  }
  void clear()
  {
    while (!q.empty())
    {
      int x=q.front(); q.pop();
      hs[x].c=hs[x].ans=0;
    }
  }
}M;
ll dfs(short x,ll y,bool z,bool kk)
{
  cnt++;
  ll ans1=M.find(z,kk,x,y);
  if (ans1!=-1) return(ans1);
  if (x==l+1)
  {
    if (kk==1) y=0;
    if (y<=m)
    {
      return(1);
    }
    return(0);
  }
  ll ans=0;
  rep(i,0,9)
  if (i==0&&kk==1) ans+=dfs(x+1,1,0,1);
  else
  {
    if (z==1)
    {
      if (i<f[x]) ans+=dfs(x+1,y*i,0,0);
      if (f[x]==i) ans+=dfs(x+1,y*i,1,0);
    } else ans+=dfs(x+1,y*i,0,0);
  }
  M.insert(z,kk,x,y,ans);
  return(ans);
}
ll js(ll x,ll y)
{
  if (x<0) return(0);
  M.clear();
  ll tmp=x; l=0;
  if (x==0) l=1,f[1]=0;
  while (tmp) l++,f[l]=tmp%10,tmp/=10;
  reverse(f+1,f+l+1);
  m=y;
  return dfs(1,1,1,1);
}
int main()
{
  ios::sync_with_stdio(false);
  ll x1,x2,y1,y2;
  cin>>x1>>x2>>y1>>y2;
  cout<<js(x2,y2)
  -js(x2,y1-1)-
  (js(x1-1,y2)-
  js(x1-1,y1-1))<<endl;
  return 0; 
}

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