给出一个 1≤N≤105 ，每次选其一个约数相除，知道得到结果为1为止，求期望次数；
期望dp，求x平均除多少次得到1；假设x有c个因子（含1和本身），E[x]表示结果； 那么E[x] = (E[1] + E[a1] + E[a2] + … + E[x] + c) / c;加c是因为每次要多走一步才能得到ai，每次选者的概率为1/c； 转换下就是E[x] = (E[1] + E[a1] + E[a2] + … + c) / (c - 1);
/***************************************** Author :Crazy_AC(JamesQi) Time :2016 File Name : *****************************************/ // #pragma comment(linker, "/STACK:1024000000,1024000000") #include <iostream> #include <algorithm> #include <iomanip> #include <sstream> #include <string> #include <stack> #include <queue> #include <deque> #include <vector> #include <map> #include <set> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <climits> using namespace std; #define MEM(x,y) memset(x, y,sizeof x) #define pk push_back #define lson rt << 1 #define rson rt << 1 | 1 #define bug cout << "BUG HERE\n" #define debug(x) cout << #x << " = " << x << endl #define ALL(v) (v).begin(), (v).end() #define lowbit(x) ((x)&(-x)) #define Unique(x) sort(ALL(x)); (x).resize(unique(ALL(x)) - (x).begin()) #define BitOne(x) __builtin_popcount(x) #define showtime printf("time = %.15f\n",clock() / (double)CLOCKS_PER_SEC) #define Rep(i, l, r) for (int i = l;i <= r;++i) #define Rrep(i, r, l) for (int i = r;i >= l;--i) typedef long long LL; typedef unsigned long long ULL; typedef pair<int,int> ii; typedef pair<ii,int> iii; const double eps = 1e-8; const double pi = 4 * atan(1); const int inf = 0x3f3f3f3f; const long long INF = 0x3f3f3f3f3f3f3f3f; const int MOD = 1e9 + 7; int nCase = 0; //精度正负、0的判断 int dcmp(double x){if (fabs(x) < eps) return 0;return x < 0?-1:1;} template<class T> inline bool read(T &n){ T x = 0, tmp = 1; char c = getchar(); while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar(); if(c == EOF) return false; if(c == '-') c = getchar(), tmp = -1; while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar(); n = x*tmp; return true; } template <class T> inline void write(T n){ if(n < 0){putchar('-');n = -n;} int len = 0,data[20]; while(n){data[len++] = n%10;n /= 10;} if(!len) data[len++] = 0; while(len--) putchar(data[len]+48); } LL QMOD(LL x, LL k) { LL res = 1LL; while(k) {if (k & 1) res = res * x % MOD;k >>= 1;x = x * x % MOD;} return res; } double dp[100010]; double dfs(int number) { if (dp[number] != -1) return dp[number]; int cnt = 2; double ans = 0; for (int i = 2;i * i <= number;++i) { if (number % i == 0) { ++cnt; ans += dfs(number / i); if (number / i != i) { ++cnt; ans += dfs(i); } } } ans += cnt; ans /= (cnt - 1); return dp[number] = ans; } int main(int argc, const char * argv[]) { // freopen("/Users/jamesqi/Desktop/in.txt","r",stdin); // freopen("/Users/jamesqi/Desktop/out.txt","w",stdout); // ios::sync_with_stdio(false); // cout.sync_with_stdio(false); // cin.sync_with_stdio(false); int kase;cin >> kase; while(kase--) { int n;cin >> n; Rep(i, 0, n) dp[i] = -1; dp[1] = 0; printf("Case %d: %.12lf\n", ++nCase, dfs(n)); } // showtime; return 0; }