题解:

每一题对的概率为min(a[i],a[i+1])/a[i]/a[i+1];

即可

代码:

#include<bits/stdc++.h>
using namespace std;
const int N=10000005;
int a[N],n,A,B,C;
int main()
{
    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;
    for (int i=1;i<n;i++)
     ans+=(double)min(a[i],a[i+1])/a[i]/a[i+1];
    printf("%.3lf",ans+(double)min(a[1],a[n])/a[1]/a[n]); 
} 

 

posted on 2017-12-24 13:38  宣毅鸣  阅读(75)  评论(0编辑  收藏  举报