poj2625 Coupons

Description Coupons in cereal boxes are numbered 1 to n, and a set of
one of each is required for a prize (a cereal box, of course). With
one coupon per box, how many boxes on average are required to make a
complete set of n coupons?

Input Input consists of a sequence of lines each containing a single
positive integer n, 1<=n<=22, giving the size of the set of coupons.

Output For each input line, output the average number of boxes
required to collect the complete set of n coupons. If the answer is an
integer number, output the number. If the answer is not integer, then
output the integer part of the answer followed by a space and then by
the proper fraction in the format shown below. The fractional part
should be irreducible. There should be no trailing spaces in any line
of ouput.

假设已经有了k个优惠券，记s=k/n
再获得一个需要t次的概率为s^(t-1) * (1-s)
所以再获得一个的期望次数为(1-s) * (s^0+2 * s^1+3 * s^2+…)
记M=s^0+2 * s^1+3 * s^2+…
则s*M=s^1+2 * s^2+3 * s^3+…
(1-s) * M=s^0+s^1+s^2+…
记上式=T
s * T=s^1+s^2+s^3+…
(1-s) * T=s^0
T=1/(1-s)
即期望次数为n/(n-k)
所以总的期望次数为n(1/n+1/(n-1)+…+1/1)
注意分数运算。

#include<cstdio>#include<cstring>#define LL long longstruct num{    LL a,b;}n1,n2;LL gcd(LL x,LL y){    return y?gcd(y,x%y):x;}LL lcm(LL x,LL y){    return x/gcd(x,y)*y;}void yf(num &nn){    LL x=gcd(nn.a,nn.b);    nn.a/=x;    nn.b/=x;}num operator + (const num n1,const num n2){    LL x=lcm(n1.b,n2.b);    num ans=(num){x/n1.b*n1.a+x/n2.b*n2.a,x};    yf(ans);    return ans;}void out(num x){    LL i,j,k,p,pp,q;    if (x.b==1)    {        printf("%lld\n",x.a);        return;    }    p=x.a/x.b;    while (p)    {        p/=10;        printf(" ");    }    printf(" ");    printf("%lld\n",x.a%x.b);    printf("%lld ",x.a/x.b);    q=x.b;    while (q)    {        q/=10;        printf("-");    }    printf("\n");    p=x.a/x.b;    while (p)    {        p/=10;        printf(" ");    }    printf(" ");    printf("%lld\n",x.b);}int main(){    LL i,j,k,m,n,p,q,x,y,z;    num ans;    while (scanf("%lld",&n)==1)    {        ans=(num){1,1};        for (i=2;i<=n;i++)          ans=ans+(num){1,i};        ans.a*=n;        yf(ans);        out(ans);    }}