233333, I originally wanted to write it myself, but halfway through the writing, I found that there was a place I couldn’t, so I went to learn it now. After Blue Queen, Baidu came to a very good blog, I thought it was simple and easy to understand, the template provided by the blogger It is also easy to remember, it is better to directly reprint qwq
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First, the Lucas theorem is given:
There are non-negative integers A, B, and prime p, A and B are written in p base as: A=a[n]a[n-1]...a[0], B=b[n]b[n -1]...b[0].
Then the combination number C(A,B) and C(a[n],b[n])×C(a[n-1],b[n-1])×...×C(a[0] ,b[0]) mod p is congruent.
That is: Lucas(n,m,p)=C(n%p,m%p)×Lucas(n/p,m/p,p), especially, Lucas(x,0,p)=1.
In fact, to put it bluntly, Lucas's theorem is to find the value of the combination number C(n, m) mod p (p is a prime number), that is (( n! /(nm)!)/m! ) mod p, and we
Also know that (a/b)mod p=a*b^(p-2)mod p (a little knowledge of inverse elements and Fermat's little theorem is used here), so we can take p in the process of calculating factorial mode, will not cause overflow. (It should be noted that the range of p handled by Lucas's theorem is roughly on the order of 10^5)
See the following question: HDU3037
Topic link: http://acm.hdu.edu.cn/showproblem.php?pid=3037
The gist of the title: m seeds should be placed on n trees, how many methods are there, and the result will be modulo the prime number p.
Analysis: m seeds can be divided into two parts: those placed on the tree and those not placed on the tree, we can imagine the n+1th tree, and think that those seeds that are not placed on the tree are placed on this imaginary tree , in this way, the question becomes how many options are there for m seeds placed on n trees. Equivalent to how many sets of non-negative integer solutions to the equation X1+X2+...+Xn+1=m, that is (X1+1)+(X2+1)+...+(Xn+1+1)=m How many positive integer solutions are there for +n+1. Baffle principle: (m+n+1) 1s, the number of solutions divided into n+1 parts==>C(n+m,n). It's obvious here, decisive Lucas.
Code:
1 #include<iostream> 2 #include<cstdio> 3 #define ll long long 4 using namespace std; 5 const int maxn=150010; 6 int t; 7 ll n,m,p; 8 ll jc[maxn]; 9 void work(ll mod) 10 { 11 jc[0]=1; 12 for(int i=1;i<=p;++i) 13 jc[i]=jc[i-1]*i%mod; 14 } 15 ll qpow(ll a,ll b) 16 { 17 ll ans=1; 18 while(b) 19 { 20 if(b&1)ans=(ans*a%p)%p; 21 a=(a*a)%p; 22 b>>=1; 23 } 24 return ans; 25 } 26 ll C(ll n,ll m) 27 { 28 if(m>n)return 0; 29 returnjc[n]*qpow(jc[m]*jc[nm]%p,p- 2 )%p; // Theorem, the inverse of a%p is equal to a^(p-2)%p 30 } 31 ll lucas(ll n,ll m) 32 { 33 if (m== 0 ) return 1 ; // c(n,0)=1; 34 else return (C(n%p,m%p)*lucas(n /p,m/p)%p)% p; 35 } 36 int main() 37 { 38 scanf( " %d " ,& t); 39 while (t-- ) 40 { 41 scanf("%lld%lld%lld",&n,&m,&p); 42 work(p); 43 printf("%lld\n",lucas(n+m,m)); 44 } 45 return 0; 46 }
For the case where p is relatively large, the factorial cannot be preprocessed and needs to be processed separately.
Topic link: http://acm.fzu.edu.cn/problem.php?pid=2020
The gist of the title: Find C(n, m) mod p. n,p are on the order of 10^9.
Code:
1 #include <iostream> 2 #include <string.h> 3 #include <stdio.h> 4 5 using namespace std; 6 typedef long long LL; 7 8 LL n,m,p; 9 10 LL quick_mod(LL a, LL b) 11 { 12 LL ans = 1; 13 a %= p; 14 while(b) 15 { 16 if(b & 1) 17 { 18 ans = ans * a % p; 19 b--; 20 } 21 b >>= 1; 22 a = a * a % p; 23 } 24 return ans; 25 } 26 27 LL C(LL n, LL m) 28 { 29 if(m > n) return 0; 30 LL ans = 1; 31 for(int i=1; i<=m; i++) 32 { 33 LL a = (n + i - m) % p; 34 LL b = i % p; 35 ans = ans * (a * quick_mod(b, p-2) % p) % p; 36 } 37 return ans; 38 } 39 40 LL Lucas(LL n, LL m) 41 { 42 if(m == 0) return 1; 43 return C(n % p, m % p) * Lucas(n / p, m / p) % p; 44 } 45 46 int main() 47 { 48 int T; 49 scanf("%d", &T); 50 while(T--) 51 { 52 scanf("%I64d%I64d%I64d", &n, &m, &p); 53 printf("%I64d\n", Lucas(n,m)); 54 } 55 return 0; 56 }