2134: 单选错位|概率与期望

相邻的两个题单独来算

相邻的两个题期望得分为1/（max（a[i],a[i+1]））

#include<cstdio>
#include<cstdlib>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int a[10000005],n,A,B,C;
double ans;
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;
a[n+1]=a[1];
for(int i=1;i<=n;i++)
{
double x=max(a[i],a[i+1]);
ans+=1/x;
}
printf("%.3lf",ans);
return 0;
}