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


[align=left]Problem Description[/align]
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?

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

2
5
17

[align=left]Sample Output[/align]

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

[align=left]Author[/align]
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==0?a:gcd(b,a%b);
}
LL lcm(LL a,LL b){
return a/gcd(a,b)*b;
}
int getLen(LL num){
int ans = 0;
while(num){
ans++;
num/=10;
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF){
LL a=1,b=1; ///a为分子,b为分母
for(int i=2;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==0){
printf("%lld\n",a/b);
}else {
LL len = getLen(res);
LL len1 = getLen(b);
LL yushu = a%b;
for(int i=0;i<=len;i++){
printf(" ");
}
printf("%lld\n%lld ",yushu,res);
for(int i=0;i<len1;i++) printf("-");
printf("\n");
for(int i=0;i<=len;i++){
printf(" ");
}
printf("%lld\n",b);
}
}
return 0;
}