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`@@ -13,10 +13,11 @@ where inv(b) = b^(p-2) and b cannot be divided by p.`
`13`	`13`	```
`14`	`14`	`这里的p非常小,分别是2和5,很容易被组合数计算表达式的分母整除。`
`15`	`15`
`16`		`-因此此题涉及到了一个非常专业的知识点,Lucas定理。见https://oi-wiki.org/math/number-theory/lucas/`
	`16`	`+因此此题涉及到了一个生僻的知识点,Lucas定理。见https://oi-wiki.org/math/number-theory/lucas/`
`17`	`17`	```
`18`	`18`	`long long Lucas(long long n, long long m, long long p) {`
`19`	`19`	`if (m == 0) return 1;`
`20`	`20`	`return (C(n % p, m % p, p) * Lucas(n / p, m / p, p)) % p;`
`21`	`21`	`}`
`22`	`22`	```
	`23`	+用这个函数,我们就直接求得了`C(n,i)%p`的值,再作为nums[i]的系数带入计算。

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