NOIP 模拟 $29\; \rm 最长不下降子序列$

题解 \(by\;zj\varphi\)

观察这个序列，发现模数很小，所以它的循环节很小。

那么可以直接在循环节上做最长上升子序列，但是循环节中的逆序对会对拼接后的答案造成影响。

没有必要找逆序对个数，直接将循环节大小个拼接在一起即可。

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            ri f(1);x=0;register char ch(gc());
            while(!isdigit(ch)) {if (ch=='-') f=0;ch=gc();}
            while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=gc();}
            return x=f?x:-x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    typedef long long ll;
    static const int N=1e5+7;
    int num[N*10],vis[N],dp[N*10],dpt[N],dpf[N],nm[N],suf[N],bg,lea,sl,len,cnt,t0,a,b,c,d;
    ll n;
    struct BIT{
        #define lowbit(x) ((x)&-(x))
        int c[N];
        inline void init() {memset(c,0,sizeof(int)*300);}
        inline void update(int x,int k) {for (ri i(x);i<=d;i+=lowbit(x)) c[i]=cmax(c[i],k);}
        inline int query(int x) {
            int res(0);
            for (ri i(x);i;i-=lowbit(i)) res=cmax(res,c[i]);
            return res;
        }
    }B;
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        cin >> n >> t0 >> a >> b >> c >> d;
        vis[num[p(cnt)]=t0]=1;
        while(1) {
            p(cnt);
            num[cnt]=(a*num[cnt-1]*num[cnt-1]+b*num[cnt-1]+c)%d;
            if (vis[num[cnt]]) {len=cnt-vis[num[cnt]];break;}
            vis[num[cnt]]=cnt;
        }
        bg=vis[num[cnt]]-1;
        if (n<=(int)1e6) {
            ri ans(0);
            for (ri i(cnt+1);i<=n;p(i)) num[i]=(a*num[i-1]*num[i-1]+b*num[i-1]+c)%d;
            for (ri i(1);i<=n;p(i)) {
                dp[i]=B.query(num[i]+1)+1;
                B.update(num[i]+1,dp[i]);
                ans=cmax(ans,dp[i]);
            }
            printf("%d\n",ans);
        } else {
            for (ri i(1);i<=len;p(i)) nm[i]=num[bg+i];
            for (ri i(2);i<=len;p(i))
                for (ri j(1);j<=len;p(j)) nm[(i-1)*len+j]=nm[j];
            ll aln=(n-bg)/len-len,ans(0);
            sl=(int)(n-(aln+len)*len-bg);
            for (ri i(1);i<=sl;p(i)) suf[i]=nm[i];
            for (ri i(1);i<=bg;p(i)) {
                dpf[i]=B.query(num[i]+1)+1;
                B.update(num[i]+1,dpf[i]);
            }
            lea=len*len;
            for (ri i(0);i<=bg;p(i)) {
                B.init();
                ri res(0),mn(INT_MAX);
                B.update(num[i]+1,dpf[i]);
                ans=cmax(ans,(ll)dpf[i]);
                for (ri j(1);j<=lea;p(j)) {
                    if (nm[j]<num[i]) continue;
                    dp[j]=B.query(nm[j]+1)+1;
                    B.update(nm[j]+1,dp[j]);
                    res=cmax(res,dp[j]);
                }
                for (ri j(lea-len+1);j<=lea;p(j))
                    if (dp[j]==res) mn=cmin(mn,nm[j]);
                B.init();
                if (res) ans=cmax(ans,aln+res),B.update(mn+1,res);
                else mn=num[i];
                for (ri j(1);j<=sl;p(j)) {
                    if (suf[j]<mn) continue;
                    dpt[j]=B.query(suf[j]+1)+1;
                    B.update(suf[j]+1,dpt[j]);
                    ans=cmax(ans,aln+dpt[j]);
                }
            }
            printf("%lld\n",ans);
        }
        return 0;
    }
}
int main() {return nanfeng::main();}