【BZOJ 2134】单选错位 概率

就是求前后两个答案一样的概率：

(1/a[i])*(1/a[i+1])*min(a[i],a[i+1])->1.0/max(a[i],a[i+1])

#include<cstdio> #include<cstring> #include<iostream> #define maxn 10000021 using namespace std; int n,a[maxn]; void pre(){ int A,B,C; 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; int main(){ pre(); for(int i=1;i<=n;i++){ ans+=1.0/max(a[i],a[i%n+1]); }printf("%.3lf",ans); return 0; }