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#include #include #include 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;}