hdu 1099 lottery

Lottery

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1086    Accepted Submission(s): 537

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

 

拿到第一張卡片的概率是 n/n ，期望次數是 1，拿到第二張的概率是 (n-1)/n， 期望次數是 n/(n-1) ...

把這些次數加起來 n/n + n/(n-1) + ... + n/1

don 分母 doneminator

num 分子 numerator

ws 空格  white space

ln  分數線 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;
}