传送门

题意：

思路： 手推了一下没想到还真的能过。
对于相邻的两个数$a_i$和$a_{i+1}$，分两种情况讨论：
(1) $a_i<=a_{i+1}$ 时，答案在$[1,a_i]$的范围内概率为$\frac{a_i}{a_{i+1}}$，正确率为$\frac{1}{a_i}$，乘起来为$\frac{1}{a_i+1}$。另一种情况正确率为$0$。
(2) $a_i>a_{i+1}$时，$a_i$在$[1,a_{i+1}]$的概率为$\frac{a_{i+1}}{a_i}$，正确率为$\frac{1}{a_{i+1}}$，乘起来为$\frac{1}{a_i}$。另一种情况正确率为$0$。
综上所述，答案为$\frac{1}{max(a_i,a_{i+1})}$。

//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid (tr[u].l+tr[u].r>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=10000100,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n,a[N],A,B,C;int main()
{
//	ios::sync_with_stdio(false);
//	cin.tie(0);scanf("%d%d%d%d%d", &n, &A, &B, &C, a + 1);for (int i = 2; i <= n; i++)a[i] = ((long long) a[i - 1] * A + B) % 100000001;for (int i = 1; i <= n; i++)a[i] = a[i] % C + 1;double ans=0.0; a[n+1]=a[1];for(int i=1;i<=n;i++) ans+=1.0/max(a[i],a[i+1]);printf("%.3f\n",ans);return 0;
}
/**/