如何才能计算机（ñ取K ）的C％M或C ++而不必调用溢出？

How can we computer (N choose K)% M in C or C++ without invoking overflow ?

有关的特殊情况下，当 N（4℃= N＆LT; = 1000）和 K（1＆LT; = K＆LT; = N）和 M = 1000000003

For the particular case when N (4<=N<=1000) and K (1<=K<=N) and M = 1000000003.

推荐答案

要计算（正选K）％M，可以分别计算出提名模数m，分母（K *（N - K ）！）模数m，然后乘以提名的分母的模反元素（米）。由于M是素数，则可以使用费马小定理来计算乘法逆。

To compute (n choose k) % M, you can separately compute the nominator (n!) modulus M and the denominator (k!*(n - k)!) modulus M and then multiply the nominator by the denominator's modular multiplicative inverse (in M). Since M is prime, you can use Fermat's Little Theorem to calculate the multiplicative inverse.

有一个很好的解释，样本code，以下链接（问题SuperSum）：

There is a nice explanation, with sample code, on the following link (problem SuperSum):

http://www.top$c$cr.com/wiki/display/tc/SRM+467