UVa 369 Combinations：如何用double算组合数

UVa 369 Combinations：如何用double算组合数2014-10-02 csdn博客 synapse7

369 - Combinations

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php？option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=305

Computing the exact number of ways that N things can be taken M at a time can be a great challenge when Nand/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

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The input to this program will be one or more lines each containing zero or more leading spaces, a value forN, 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

 1006205186 00

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.

注意这句话：

“You may assume that the final value of C will fit in a 32-bit Pascal LongInt or a C long.”

完整代码：

/*0.022s*/ #include<cstdio> int main（）{int N, M, i;double C;while （scanf（"%d%d", &N, &M）, N）{C = 1.0;for （i = N; i > N - M; --i） C *= （double）i;for （i = 2; i <= M; ++i） C /= （double）i;printf（"%d things taken %d at a time is %.0f exactly.
", N, M, C）;}return 0;}