UVa10288

Coupons in cereal boxes are numbered 1 to n, and a set of one of each is required for a prize (a cerealbox, of course). With one coupon per box, how many boxes on average are required to make a completeset of n coupons?InputInput consists of a sequence of lines each containing a single positive integer n, 1 ≤ n ≤ 33, giving thesize of the set of coupons. Input is terminated by end of file.OutputFor each input line, output the average number of boxes required to collect the complete set of ncoupons. If the answer is an integer number, output the number. If the answer is not integer, thenoutput the integer part of the answer followed by a space and then by the proper fraction in the formatshown below. The fractional part should be irreducible. There should be no trailing spaces in any lineof output.Sample Input2517Sample Output3   511 --   12   34046358 ------   720720

假设当前已经拿到了K种奖券
从箱子中取出一种新奖券的概率是(n-K)/n
平均要取n/(n-K)次
所以ans=∑n−1i=0n/(n−i)

#include<iostream>#include<cstdlib>#include<cstdio>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<ctime>#include <string.h>#include<math.h>using namespace std;#define ll long long#define pii pair<int,int>const int inf=1e9+7;struct Num{    ll fz,fm;    ll gcd(ll a,ll b)const{        return b?gcd(b,a%b):a;    }    Num(){        fz=0,fm=1;    }    Num(ll fz,ll fm){        this->fz=fz;        this->fm=fm;    }    Num operator+(const Num&x)const{        Num ans;        ans.fm=fm*x.fm;        ans.fz=fz*x.fm+fm*x.fz;        ll d=gcd(ans.fz,ans.fm);        ans.fz/=d;        ans.fm/=d;        return ans;    }};int len(ll x){    int len=0;    while(x){        x/=10;        ++len;    }    return len;}void print(const Num&x){    if(x.fm==1){        printf("%lld\n",x.fz);    }    else{        ll ansInt=x.fz/x.fm;        int ansIntLen=len(ansInt);        Num ansFloat=x;        ansFloat.fz-=x.fm*ansInt;        int fmLen=len(x.fm);        for(int i=0;i<=ansIntLen;++i){            putchar(' ');        }        printf("%lld\n",ansFloat.fz);        printf("%lld ",ansInt);        for(int i=0;i<fmLen;++i){            putchar('-');        }        putchar('\n');        for(int i=0;i<=ansIntLen;++i){            putchar(' ');        }        printf("\%lld\n",ansFloat.fm);    }}int main(){    //freopen("/home/lu/Documents/r.txt","r",stdin);    int n;    while(~scanf("%d",&n)){        Num ans;        for(int i=0;i<n;++i){            ans=ans+Num(n,n-i);        }        print(ans);    }    return 0;}