https://www.luogu.com.cn/problem/P1297
考虑期望的线性性。E(x + y + z) = E(x) + E(y) + E(z)
跟x,y,z独立性无关`
E(i):表示第i个题正确的期望,这里也就是概率 * 1,所以跟概率是相等的。
E(i) = min(a[i-1),a[i])/(a[i-1] * a[i])
(a[i-1] * a[i]): 表示所有的情况数
min(a[i-1),a[i]):表示两个题答案一一对应的情况数
#include <stdio.h> #include <cstring> #include <algorithm> #include <vector> #include <stack> #include <queue> #include <iostream> #include <map> #define go(i, l, r) for(int i = (l), i##end = (int)(r); i <= i##end; ++i) #define god(i, r, l) for(int i = (r), i##end = (int)(l); i >= i##end; --i) #define ios ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0) #define debug_in freopen("in.txt","r",stdin) #define debug_out freopen("out.txt","w",stdout); #define pb push_back #define all(x) x.begin(),x.end() #define fs first #define sc second using namespace std; typedef long long ll; typedef unsigned long long ull; typedef pair<ll,ll> pii; const ll maxn = 1e7+10; const ll maxM = 1e6+10; const ll inf_int = 1e8; const ll inf_ll = 1e17; template<class T>void read(T &x){ T s=0,w=1; char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();} while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar(); x = s*w; } template<class H, class... T> void read(H& h, T&... t) { read(h); read(t...); } void pt(){ cout<<'\n';} template<class H, class ... T> void pt(H h,T... t){ cout<<" "<<h; pt(t...);} //-------------------------------------------- int n,A,B,C; int a[maxn]; void solve(){ double ans = 0; a[n+1] = a[1]; go(i,2,n+1){ ans += min(a[i-1],a[i])/((double)a[i-1] * a[i]); } printf("%.3f\n",ans); } int main() { // debug_in; // debug_out; 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; solve(); return 0; }