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 ==  ? a : gcd(b, a % b);
 }

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

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

     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 == ) {
             denominator = ;
             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 = ;
     while (n) {
         n /= ;
         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)) {
         printf("%lld\n", b);
     } else {
         a = b / c;
         m = countLen(a);
         b -= a * c;
         n = countLen(c);
         for (int i = ; i <= m; i++) {
             printf(" ");
         }
         printf("%lld\n%lld ", b, a);
         for (int i = ; i < n; i++) {
             printf("-");
         }
         puts("");
         for (int i = ; i <= m; i++) {
             printf(" ");
         }
         printf("%lld\n", c);
     }
 }

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

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

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

     return ;
 }