长春商城网站建设/如何让百度搜索到自己的网站

https://vjudge.net/problem/UVA-10288

大街上到处在卖彩票，一元钱一张。购买撕开它上面的锡箔，你会看到一个漂亮的图案。

图案有n种，如果你收集到所有n（n≤33）种彩票，就可以得大奖。

请问，在平均情况下，需要买多少张彩票才能得到大奖呢？

答案以带分数形式输出

 

例：当n=5时

思路简单，就是输出麻烦

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn=500;
struct fraction
{long long numerator;long long denominator;fraction(){numerator=0;denominator=1;}fraction(long long num){numerator=num;denominator=1;}fraction(long long a,long long b){numerator=a;denominator=b;this->reduction();}void operator =(const long long num){numerator=num;denominator=1;this->reduction();}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);fraction tmp=fraction(numerator*(b.denominator/gcdnum)+b.numerator*(denominator/gcdnum),denominator/gcdnum*b.denominator);tmp.reduction();return tmp;}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{fraction tmp=fraction(numerator*b.numerator,denominator*b.denominator);tmp.reduction();return tmp;}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(numerator,denominator);numerator/=gcdnum;denominator/=gcdnum;}void print(){if(denominator==1) { printf("%lld\n",numerator); }else{long long num=numerator/denominator;long long tmp=num;int len=0;while(tmp){len++;tmp/=10;}for(int i=1;i<=len;i++) printf(" ");if(len) printf(" ");printf("%lld\n",numerator%denominator);if(num) printf("%lld ",num);tmp=denominator;while(tmp){printf("-");  tmp/=10;}puts("");for(int i=1;i<=len;i++) printf(" ");if(len) printf(" ");printf("%lld\n",denominator);}}
}f[maxn];
int main()
{int n;while(scanf("%d",&n)!=EOF){f[0]=0;for(int i=1;i<=n;i++) f[i]=f[i-1]+fraction(1,i);f[n]=f[n]*n;f[n].print();}return 0;
}