BZOJ 2134 单选错位 ——期望DP

发现概率是∑1/两道题答案相同的概率，

稍加化简

#include <map>
#include <ctime>
#include <cmath>
#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (int i=j;i<=k;++i)
#define D(i,j,k) for (int i=j;i>=k;--i)

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;
    F(i,1,n-1) ans+=1.0/max(a[i],a[i+1]);
    ans+=1.0/max(a[n],a[1]);
    printf("%.3f\n",ans);
}