Description
Computing
the exact number of ways that N things can be taken M at a time can be
a great challenge when N and/or M become very large. Challenges are the
stuff of contests. Therefore, you are to make just such a computation
given the following:
GIVEN: 5 <= N <= 100; 5 <= M <= 100; M <= N
Compute the EXACT value of: C = N! / (N-M)!M!
You may assume that the final value of C will fit in a 32-bit
Pascal LongInt or a C long. For the record, the exact value of 100! is:
93,326,215,443,944,152,681,699,238,856,266,700,490,715,968,264,381,621,
468,592,963,895,217,599,993,229,915,608,941,463,976,156,518,286,253,
697,920,827,223,758,251,185,210,916,864,000,000,000,000,000,000,000,000
Input
The
input to this program will be one or more lines each containing zero or
more leading spaces, a value for N, one or more spaces, and a value for
M. The last line of the input file will contain a dummy N, M pair with
both values equal to zero. Your program should terminate when this line
is read.
Output
The output from this program should be in the form:
N things taken M at a time is C exactly.
Sample Input
100 6
20 5
18 6
0 0
Sample Output
100 things taken 6 at a time is 1192052400 exactly.
20 things taken 5 at a time is 15504 exactly.
18 things taken 6 at a time is 18564 exactly.
Source
Solution:
more..
import java.math.BigInteger;
import java.util.*;
public class P1306 {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
while(true){
String temp = sc.nextLine();
StringTokenizer st = new StringTokenizer(temp, " ");
String sn = st.nextToken();
String sm = st.nextToken();
int n =Integer.parseInt(sn);
int m = Integer.parseInt(sm);
if(n==0){
break;
}
BigInteger bn = new BigInteger(sn);
BigInteger bm = new BigInteger(sm);
BigInteger answer;
answer = (factorial(bn).divide(((factorial(bn.subtract(bm)).multiply(factorial(bm))))));
System.out.println(bn+" things taken "+bm+" at a time is "+answer+" exactly.");
}
}
public static BigInteger factorial( BigInteger n )
{
if( n.compareTo(new BigInteger("1"))<=0 ) // base case
return new BigInteger("1");
else
return n.multiply(factorial( n.subtract(new BigInteger("1"))));
}
}
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