hdu 1099 Lottery
題意：1~n編號的彩票，要買全，等概率條件下平均要買幾張。
已經買了m張時，買中剩下的概率為1-m/n，則要買的張數為1/(1-m/n)
n=2，s=1+1/(1-1/2);n=3，s=1+1/(1-1/3)+1/(1-2/3)
s=1+1/(1-1/n)+1/(1-2/n)+1/(1-3/n)+……+1/(1-(n-1)/n)=n/n+n/(n-1)+n/(n-2)+……+n/1=sum(n/i),i=1~n
b/a+d/c=(bc+ad)/(ac)
然後遞推著通分，化簡；輸出。

///實現 sum(n/i) (i=1~n)
#include<iostream>
#include<cstdio>
#include
 
<cmath>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
using namespace std;
#define LL long long
LL gcd(LL a,LL b)
{
    return b==0?a:gcd(b,a%b);
}
struct fen{
    LL a,b;

};
fen& add(struct fen&a,struct fen&b)
{
    a.a=a.a*b.b+a.b*b.a;
    a.b
 
=a.b*b.b;
    LL t=gcd(a.a,a.b);
    a.a/=t;
    a.b/=t;
    return a;
}
int ditNum(LL a)
{
    int cnt=0;
    while(a)
    {
        cnt++;
        a/=10;
    }
    return cnt;
}
int main()
{
    int n;
    LL fenz=0,fenm=0;
    struct fen f,f1;
    while(~scanf("%d",&n))
    {
        f.a=n;f.b=1;
        
 
for(int i=2;i<=n;i++)
        {
            f1.a=n;f1.b=i;
            f=add(f,f1);
        }
        if(f.a%f.b==0)
        {
            cout<<f.a/f.b<<endl;
        }
        else
        {
            LL t=f.a/f.b;
            int a=ditNum(t),b=ditNum(f.b);
            a++;
            int x=a;
            while(a--)
                cout<<" ";
            cout<<f.a-t*f.b<<endl;
            cout<<t<<" ";
            while(b--)
                cout<<"-";
            cout<<endl;
            while(x--)
                cout<<" ";
            cout<<f.b<<endl;
        }
    }
}

HDU1099---數學 | 思維