题目链接：
https://cn.vjudge.net/problem/UVA-10288
题目大意：
一种刮刮卡一共有n种图案，每张可刮出一个图案，收集n种就有奖，问平均情况下买多少张才能中奖？用最简的分数形式表示答案。n<=33。
解题思路：
假设现在已刮到k个图案了，刮到新图案的概率是(n-k)/n，即若要再收集一个新图案平均要刮s=n/(n-k)次。所以只需要穷举k=1 to n，累加s的和就行了。注意式子可以将分子n提取出来。
先附上分数类模板
1 struct Fraction//分数类 2 { 3 ll num, den;//num为分子，den为分母 4 Fraction(ll num = 0, ll den = 1) 5 { 6 if(den < 0) 7 { 8 num = -num; 9 den = -den; 10 } 11 assert(den != 0);//den=0程序终止 12 ll g = __gcd(abs(num), den); 13 this->num = num / g; 14 this->den = den / g; 15 } 16 Fraction operator + (const Fraction& o)const 17 { 18 return Fraction(num * o.den + den * o.num, den * o.den); 19 } 20 Fraction operator - (const Fraction& o)const 21 { 22 return Fraction(num * o.den - den * o.num, den * o.den); 23 } 24 Fraction operator * (const Fraction& o)const 25 { 26 return Fraction(num * o.num, den * o.den); 27 } 28 Fraction operator / (const Fraction& o)const 29 { 30 return Fraction(num * o.den, den * o.num); 31 } 32 bool operator < (const Fraction& o)const 33 { 34 return num * o.den < den * o.num; 35 } 36 bool operator == (const Fraction& o)const 37 { 38 return num * o.den == den * o.num; 39 } 40 };
注意输出格式（很伤）
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 struct Fraction//分数类 5 { 6 ll num, den;//num为分子，den为分母 7 Fraction(ll num = 0, ll den = 1) 8 { 9 if(den < 0) 10 { 11 num = -num; 12 den = -den; 13 } 14 assert(den != 0);//den=0程序终止 15 ll g = __gcd(abs(num), den); 16 this->num = num / g; 17 this->den = den / g; 18 } 19 Fraction operator + (const Fraction& o)const 20 { 21 return Fraction(num * o.den + den * o.num, den * o.den); 22 } 23 Fraction operator - (const Fraction& o)const 24 { 25 return Fraction(num * o.den - den * o.num, den * o.den); 26 } 27 Fraction operator * (const Fraction& o)const 28 { 29 return Fraction(num * o.num, den * o.den); 30 } 31 Fraction operator / (const Fraction& o)const 32 { 33 return Fraction(num * o.den, den * o.num); 34 } 35 bool operator < (const Fraction& o)const 36 { 37 return num * o.den < den * o.num; 38 } 39 bool operator == (const Fraction& o)const 40 { 41 return num * o.den == den * o.num; 42 } 43 }; 44 int main() 45 { 46 ll n, k, ans, cases = 0; 47 while(cin >> n) 48 { 49 Fraction ans; 50 for(int i = 1; i <= n; i++) 51 { 52 ans = ans + Fraction(n, i); 53 } 54 ll t = ans.num / ans.den; 55 ans.num -= t * ans.den; 56 if(ans.num == 0) 57 { 58 cout<<t<<endl; 59 } 60 else 61 { 62 string s; 63 stringstream ss, ss1; 64 ss << t; 65 ss >> s; 66 int kongge = s.size(); 67 for(int i = 0; i <= kongge; i++)cout<<" "; 68 cout<<ans.num<<" "; 69 70 ss1 << ans.den; 71 ss1 >> s; 72 int duanxian = s.size(); 73 cout<<t<<" "; 74 for(int i = 0; i < duanxian; i++)cout<<"-"; 75 cout<<" "; 76 for(int i = 0; i <= kongge; i++)cout<<" "; 77 cout<<ans.den<<" "; 78 } 79 } 80 return 0; 81 }