hdu 1099(数学)

Lottery

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3333    Accepted Submission(s): 1489

Problem Description

Eddy's
company publishes a kind of lottery.This set of lottery which are
numbered 1 to n, and a set of one of each is required for a prize .With
one number per lottery, how many lottery 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 lottery 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.

 

Sample Input

2
5
17

 

Sample Output

3
5
11 --
12
340463
58 ------
720720

 

Author

eddy

题意很难理解。。看了别人的翻译才懂。。

题目的大概意思是说一套彩票有编号1到n共n种，张数不限，问你平均买多少张能把编号为1到n的n中彩票全买下来，也就是求期望。

也就是求n/n+n/(n-1)+...+n/1..迭代求解 a/b+1/i = ai+b/(bi)

第一组测试用例后面明明有一个空格...结果加了之后还报了一次格式错误。。。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b){
    return b==?a:gcd(b,a%b);
}
LL lcm(LL a,LL b){
    return a/gcd(a,b)*b;
}
int getLen(LL num){
    int ans = ;
    while(num){
        ans++;
        num/=;
    }
    return ans;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        LL a=,b=; ///a为分子,b为分母
        for(int i=;i<=n;i++){
            a = a*i+b;
            b = b*i;
            LL d = gcd(a,b);
            a/=d;
            b/=d;
        }
        a=a*n;
        LL d = gcd(a,b);
        a/=d,b/=d;
        LL res = a/b;
        if(a%b==){
            printf("%lld\n",a/b);
        }else {
            LL len = getLen(res);
            LL len1 = getLen(b);
            LL yushu = a%b;
            for(int i=;i<=len;i++){
                printf(" ");
            }
            printf("%lld\n%lld ",yushu,res);
            for(int i=;i<len1;i++) printf("-");
            printf("\n");
             for(int i=;i<=len;i++){
                printf(" ");
            }
            printf("%lld\n",b);
        }
    }
    return ;
}