Combinations

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:

Compute the EXACT value of:

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 and Output

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.

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.

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作法 : 遞迴建表, (PS. 只有加法)

#include<stdio.h>
int main() {
	int N, M, i, j;
	long long int PASCAL[101][101] = {0};
	PASCAL[1][0] = PASCAL[1][1] = 1;
	for(i = 2; i <= 100; i++) {
		PASCAL[i][0] = 1;
		for(j = 1; j < i; j++)
			PASCAL[i][j] = PASCAL[i-1][j] + PASCAL[i-1][j-1];
		PASCAL[i][i] = 1;
	}
	while(scanf("%d %d", &N, &M) == 2) {
		if(N == 0 && M == 0)
			break;
		printf("%d things taken %d at a time is %lld exactly.\n", N, M, PASCAL[N][M]);
	}
    return 0;
}

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