题意
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;
思路
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;
}