uva10288 Coupons 【概率 分数】

题目：

题意：

一共n种不同的礼券，每次得到每种礼券的概率相同。求期望多少次可以得到所有n种礼券。结果以带分数形式输出。1<= n <=33.

思路：

假设当前已经得到k种，获得新的一种的概率是(n-k)/n，则对应期望是n/(n-k)。求和得到步数期望是n/n+n/(n-1)+...+n/1=n*sum(1/i) (1<= i <= n)。需要注意及时约分，用分数类模板。

程序：

#include <cstdio>
#include <cassert>
#include <cstdlib>

long long gcd (long long a, long long b) {
    return b == 0 ? a : gcd(b, a % b);
}

class fraction {
public:
    fraction() {
        numerator = 0;
        denominator = 1;
    }
    fraction(long long num) {
        numerator = num;
        denominator = 1;
    }
    fraction (long long a, long long b) {
        assert(b != 0);
        if (b < 0) {
            numerator = -a;
            denominator = -b;
        } else {
            numerator = a;
            denominator = b;
        }
        this->reduction();
    }

    void operator = (long long num) {
        numerator = num;
        denominator = 1;
    }

    void operator = (const fraction &b) {
        numerator = b.numerator;
        denominator = b.denominator;
        this->reduction();
    }

    fraction operator + (const fraction &b) const {
        long long gcdnum = gcd(denominator, b.denominator);
        return fraction(numerator*(b.denominator/gcdnum) + b.numerator*(denominator/gcdnum), denominator/gcdnum*b.denominator);
    }

    fraction operator + (const int b) const {
        return ((*this) + fraction(b));
    }

    fraction operator - (const fraction &b) const {
        return ((*this) + fraction(-b.numerator, b.denominator));
    }

    fraction operator - (const int &b) const {
        return ((*this) - fraction(b));
    }

    fraction operator * (const fraction &b) const {
        return fraction(numerator*b.numerator, denominator * b.denominator);
    }

    fraction operator * (const int &b) const {
        return ((*this) * fraction(b));
    }

    fraction operator / (const fraction &b) const {
        return ((*this) * fraction(b.denominator, b.numerator));
    }

    void reduction() {
        if (numerator == 0) {
            denominator = 1;
            return;
        }
        long long gcdnum = gcd(abs(numerator), denominator);
        numerator /= gcdnum;
        denominator /= gcdnum;
    }

public:
    long long numerator;//分子
    long long denominator;//分母
};

int countLen (long long n) {
    int m = 0;
    while (n) {
        n /= 10;
        m++;
    }
    return m;
}

void printF (const fraction &f){
    long long a, b, c;
    int m, n;
    b = f.numerator;
    c = f.denominator;
    if (c == (long long)1) {
        printf("%lld
", b);
    } else {
        a = b / c;
        m = countLen(a);
        b -= a * c;
        n = countLen(c);
        for (int i = 0; i <= m; i++) {
            printf(" ");
        }
        printf("%lld
%lld ", b, a);
        for (int i = 0; i < n; i++) {
            printf("-");
        }
        puts("");
        for (int i = 0; i <= m; i++) {
            printf(" ");
        }
        printf("%lld
", c);
    }
}

int main() {
    int n, maxn = 33;
    fraction f[maxn + 1];

    for (int i = 1; i <= maxn; i++) {
        f[i] = f[i - 1] + fraction(1, i);
    }

    while (~scanf("%d", &n)) {
        printF(f[n] * n);
    }

    return 0;
}