[POJ2625][UVA10288]Coupons

Description

Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cereal box, of course). With one coupon per box, how many boxes on average are required to make a complete set of n coupons?

Input

Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=22, giving the size of the set of coupons.

Output

For each input line, output the average number of boxes required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of ouput.

题目分析

这是一道求期望的题，假设当前已经有 \(k\) 种 Coupons， 那么获得新的 Coupons 的概率是 \((n-k)/n\)，所以需要步数的期望是 $n/(n-k) $。求和得到步数的期望是 \(n/n + n/(n-1) + \cdots + n/1 = n\sum_{i=1}^{n} 1/i\)

这道题可以递推来做，但是我懒。（还有，luogu上的难度是 省选- 你怕是在逗我吧。）

#include <cstdio>
#include <iostream>
#include <cstdlib>

using namespace std;

typedef long long qword;

const qword fen[] = {-1,1,2,6,12,60,60,420,840,2520,2520,27720,27720,360360,360360,360360,720720,12252240,12252240,232792560,232792560,232792560,232792560,5354228880,5354228880,26771144400,26771144400,80313433200,80313433200,2329089562800,2329089562800,72201776446800,144403552893600,144403552893600,};

inline qword __gcd(qword a, qword b) {
    if (a == 1 || b == 1) return 1;
    if (b == 0) return a;
    else return __gcd(b, a % b);
}
int getdig(qword x) {
    int res = 0;
    while (x) {
        res ++;
        x /= 10;
    }
    return res;
}

inline void work(int n) {
    if (n == 1) {cout << 1 << endl; return;}
    qword fenzi = 0, fenmu = fen[n];
    qword ans = 0;
    for (qword i = 1; i <= n; ++ i) {
        if (i < n) fenzi += fenmu / i;
        if (fenzi >= fenmu) {ans += (fenzi / fenmu); fenzi %= fenmu;}
        if (i == n) {fenmu /= n;ans *= n; ans += 1;}
        if (fenzi >= fenmu) {ans += (fenzi / fenmu); fenzi %= fenmu;}
    }
    int uuu = __gcd(fenzi, fenmu);
    fenzi /= uuu;
    fenmu /= uuu;

    if (fenzi == 0) {
        cout << ans << endl;
    } else {
        for (int i = 1; i <= getdig(ans) + 1; ++ i) printf(" ");
        printf("%I64d\n%I64d ", fenzi, ans);
        for (int i = 1; i <= getdig(fenmu); ++ i) printf("-");
        putchar('\n');
        for (int i = 1; i <= getdig(ans) + 1; ++ i) printf(" ");
        printf("%I64d\n", fenmu);
    }
}

int main() {
    int n;
    while(cin >> n)
    work(n);
}