HDUOJ 1099——Lottery

Lottery

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2061 Accepted Submission(s): 941

[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

[align=left]Recommend[/align]
JGShining

sum=n*∑(1/i);----->n*(1+1/2+1/3+.....+1/n);

题意不好懂.....表示看来白天没明白....之后看了别人的讲的题意才懂的.....

之后的就不难了！

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
using namespace std;
int gcd(_int64 a,_int64 b)
{
if(b==0)
return a;
gcd(b,a%b);
};

void swap(_int64 *a,_int64 *b)
{
*a^=*b,
*b^=*a,
*a^=*b;
}
int main()
{
int n,i;
char str[20]={'\0'},num[20]={'\0'};
_int64 a,b,c,real;
while(scanf("%d",&n)!=EOF)
{
a=b=1;
for(i=2;i<=n;i++)
{
a=a*i+b;
b*=i;
if(a<b) swap(a,b);
c=gcd(a,b);
a/=c;
b/=c;
}
if((n*a)%b)
{
real=(a/b)*n;
a%=b;
a*=n;
real+=a/b;
a%=b;
c=gcd(a,b);
a/=c;
b/=c;
itoa(b,str,10);
itoa(real,num,10);
for(i=0;i<=strlen(num);i++)
printf(" ");
printf("%I64d\n%I64d ",a,real);
for(i=0;i<strlen(str);i++)
printf("-");
puts("");
for(i=0;i<=strlen(num);i++)
printf(" ");
printf("%I64d\n",b);
}
else
printf("%I64d\n",n*a/b);
}
return 0;
}

View Code