UVA 10288 Coupons

【解题方法】假设当前已经有k种不同的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)。求和可得到答案为nsigma(1/i)！

题目要求用分数表示，那么加个分数类重载就OK了！

【代码君】

#include <assert.h>
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
long long gcd(long long a, long long b)
{
return b == 0 ? a : gcd(b, a%b);
}
struct Fraction
{
long long num;
long long den;
Fraction(long long num = 0, long long den = 1)
{
if(den < 0)
{
num = -num;
den = -den;
}
assert(den != 0);
long long g = gcd(abs(num), den);
this->num = num / g;
this->den = den / g;
}
Fraction operator +(const Fraction &o) const
{
return Fraction(num * o.den + den * o.num, den * o.den);
}
Fraction operator -(const Fraction &o) const
{
return Fraction(num * o.den - den * o.num, den * o.den);
}
Fraction operator *(const Fraction &o) const
{
return Fraction(num * o.num, den * o.den);
}
Fraction operator /(const Fraction &o) const
{
return Fraction(num * o.den, den * o.num);
}
bool operator <(const Fraction &o) const
{
return num * o.den < den * o.num;
}
bool operator ==(const Fraction &o) const
{
return num * o.den == den * o.num;
}
};
int solve(long long x)
{
int t = 0;
while(x)
{
t++;
x = x/10;
}
return t;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
Fraction ans(n,1);
Fraction tmp(0,1);
for(int i=1; i<=n; i++)
tmp=tmp+Fraction(1,i);
ans=ans*tmp;
long long a,b,c;
a=ans.num;
b=ans.den;
if(a/b!=0)
{
long long gd=gcd(a,b);
a/=gd;
b/=gd;
c=a/b;
a=a-c*b;
}
else
c=0;
if(a==0)
printf("%lld\n",c);
else
{
int l1,l2,l3;
long long t1=c,t2=a,t3=b;
l1=0;
while(t1)
{
t1/=10;
l1++;
}
l2=0;
while(t2)
{
t2/=10;
l2++;
}
l3=0;
while(t3)
{
t3/=10;
l3++;
}
if(l1!=0)
{
for(int i=0; i<=l1; i++) printf(" ");
printf("%lld\n",a);
printf("%lld ",c);
int len=max(l2,l3);
for(int i=0; i<len; i++) printf("-");
printf("\n");
for(int i=0; i<=l1; i++) printf(" ");
printf("%lld\n",b);
}
else
{
printf("%I64d\n",a);
int len=max(l2,l3);
for(int i=0; i<len; i++)
printf("-");
printf("\n");
printf("%I64d\n",b);

}
}
}
return 0;
}