hdu 1099 lottery

Lottery

[b]Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 1086 Accepted Submission(s): 537

[/b]

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

Recommend

JGShining

拿到第一张卡片的概率是 n/n ，期望次数是 1，拿到第二张的概率是 (n-1)/n， 期望次数是 n/(n-1) ...

把这些次数加起来 n/n + n/(n-1) + ... + n/1

don 分母 doneminator

num 分子 numerator

ws 空格 white space

ln 分数线 line

http://acm.hdu.edu.cn/forum/read.php?tid=16311&keyword=1099

#include <cstdlib>
#include <cstdio>
#include <cassert>
#include <cstring>
#include <algorithm>

long long GCD (long long a, long long b) {
if (b == 0) {
return a;
}
return GCD (b, a % b);
}

int main () {
long long n;
while (scanf ("%lld", &n) == 1) {
long long don = 1, num = 0;
for (long long i=1; i<=n; ++i) {
num = num * i + n * don;
don = don * i;
long long gcd = GCD (don, num);
don = don / gcd;
num = num / gcd;
}
if (num % don == 0) {
printf ("%lld\n", num / don);
} else {
static char strInt[1024], strNum[1024], strDon[1024];
static char strWS[1024], strLn[1024];
static char ws[] = "                              ";
static char ln[] = "------------------------------";

sprintf (strInt, "%lld", num / don);
sprintf (strNum, "%lld", num % don);
sprintf (strDon, "%lld", don);

memset (strWS, 0, sizeof (strWS));
strncpy (strWS, ws, strlen (strInt));

memset (strLn, 0, sizeof (strLn));
strncpy (strLn, ln, std::max (strlen (strNum), strlen (strDon)));

printf ("%s %s\n%s %s\n%s %s\n", strWS, strNum, strInt, strLn, strWS, strDon);
}
}
return 0;
}