BZOJ3613: [Heoi2014] South Park is full of light flakes

analyze:

It takes a long time to construct the data, you can use Qin Jiushao to optimize it.

Dichotomous answer + greedy, that is: each b[i] is as small as possible and meets the meaning of the question. During the enumeration process, determine whether there is a b[i-1]>a[i]+x

If it exists, look to the right

If it doesn't exist, look left

Attached code:

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <queue>
//#include <ctime>
using namespace std;
#define N 5000005
#define ll long long
int a [N], f [N], b [N], n, Sa, Sb, Sc, Sd, mod;
int F(int x)
{
	if(f[x]) return f[x];
	return f[x]=((1ll*Sa*a[x]%mod+Sb)*a[x]*a[x]%mod+(1ll*Sc*a[x]%mod+Sd)%mod)%mod;
}
int check(int x)
{
	b[0]=-1<<30;
	for(int i=1;i<=n;i++)
	{
		if(a[i]>b[i-1]+x)
		{
			b[i]=a[i]-x;
		}else
		{
			if(a[i]+x<b[i-1])return 0;
			b[i]=b[i-1];
		}
	}
	return 1;
}
intmain()
{
	a[0]=0;
	scanf("%d%d%d%d%d%d%d",&n,&Sa,&Sb,&Sc,&Sd,&a[1],&mod);
	for(int i=2;i<=n;i++)a[i]=(F(i-1)+F(i-2))%mod;
	int l=0,r=1<<30;
	while(l<r)
	{
		int m=(l+r)>>1;
		if(check(m)) r=m;
		else l=m+1;
	}
	printf("%d\n",l);
	return 0;
}

　　Stuck a constant (you need to tap a constant on loj, bzoj and Luogu don’t need it, they run fast with O2)