Preprocess the binomial coefficient mod a prime number MOD with time cost of O(n)_算法分析与设计题目当算a^n modm,a>1,n>0的正整数,n非常大时,a的n次方过大如何计-优快云博客

First remember that $\binom{n}{m}=\frac{n!}{m!(n-m)!}$ , and all we have to do is to calculate all the n fatorials, namely Factorial[i]=i!, 0<=i<=n and their corresponding inverses Factorial_Inverse[i] in O(n) time. And when we want to calculate certain binomial coefficient, we simply need to multiple three numbers that have been preprocessed by us, namely, Factorial[n], Factorial_Inverse[m], Factorial_Inverse[n-m].

For Factorial[i], it is quite trivial.

Factorial[0]=1;

for(int i=0;i<=n;i++)Factorial[i]=(Factorial[i-1]*i)%MOD;

For Factorial_Inverse[i], it is a bit more complicated. if we tend to get the inverse in linear time, we have to kind of use the information that has been calculated several steps before. Thus we only need to traverse the whole n-length long array once, with time cost if O(n).

When we want to calculate Factorial_Inverse[i], we want to use the information of Factorial_Inverse[1...i-1]. Thus we do the following observation. Let t=MOD/i, k=MOD%i, and t*i+k $\equiv$ 0 (mod MOD). Multiple both sides of this equation by Factorial_Inverse[i] and Factorial_Inverse[k], we get t*Factorial_Inverse[k]+Factorial_Inverse[i] $\equiv$ 0 (mod MOD).

Conclusion Factorial_Inverse[i] $\equiv$ -MOD/i*Factorial_Inverse[MOD%i] (mod MOD), with MOD%i<i. And see how we used the information calculated before.

Rewrite this equation we have Factorial_Inverse[i]=(MOD-MOD/i*Factorial_Inverse[MOD%i])%MOD.

And to save more time, we continue to preprocess Factorial_Inverse[i] by multiplying the previous i terms together.

Factorial[1]=1;

for(int i=2;i<=n;i++)Factorial_Inverse[i]=(MOD-MOD/i*Factorial_Inverse[MOD%i])%MOD;

for(int i=2;i<=n;i++)Factorial_Inverse[i]=(Factorial_Inverse[i]*Factorial_Inverse[i-1])%MOD;