网上证明很多，虽然没看懂。。。。
主要解决大组合数取模的情况
费马小定理求大组合数：
a^(p-1)=1%p;
两边同除a
a^(p-2)=1/a%p;
C（n,m）= n!/(m!*(n-m)!)
所以C（n,m）=f(n)*qpow(f(m)*f(n-m),MOD-2))%MOD
预处理组合数f
百度之星2016 1003
先推公式，再lucas
p很大的情况 1e9+7
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #include<cstdlib> 5 #include<cstdio> 6 #include<set> 7 #include<map> 8 #include<vector> 9 #include<cstring> 10 #include<stack> 11 #include<cmath> 12 #include<queue> 13 #define clc(a,b) memset(a,b,sizeof(a)) 14 #include <bits/stdc++.h> 15 const int maxn = 20005; 16 const int inf=0x3f3f3f3f; 17 const double pi=acos(-1); 18 typedef long long LL; 19 using namespace std; 20 const LL MOD = 1e9+7; 21 22 LL exp_mod(LL a, LL b, LL p) 23 { 24 LL res = 1; 25 while(b != 0) 26 { 27 if(b&1) res = (res * a) % p; 28 a = (a*a) % p; 29 b >>= 1; 30 } 31 return res; 32 } 33 34 LL Comb(LL a, LL b, LL p) 35 { 36 if(a < b) return 0; 37 if(a == b) return 1; 38 if(b > a - b) b = a - b; 39 40 LL ans = 1, ca = 1, cb = 1; 41 for(LL i = 0; i < b; ++i) 42 { 43 ca = (ca * (a - i))%p; 44 cb = (cb * (b - i))%p; 45 } 46 ans = (ca*exp_mod(cb, p - 2, p)) % p; 47 return ans; 48 } 49 50 LL Lucas(int n, int m, int p) 51 { 52 LL ans = 1; 53 54 while(n&&m&&ans) 55 { 56 ans = (ans*Comb(n%p, m%p, p)) % p; 57 n /= p; 58 m /= p; 59 } 60 return ans; 61 } 62 63 int main() 64 { 65 int n, m; 66 LL p; 67 while(~scanf("%d%d", &n, &m)) 68 { 69 p=MOD; 70 printf("%I64d ", Lucas(n+m-4, m-2, p)); 71 } 72 return 0; 73 }
p在100000左右
HDU 3037
1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #include<cstdlib> 5 #include<cstdio> 6 #include<set> 7 #include<map> 8 #include<vector> 9 #include<cstring> 10 #include<stack> 11 #include<cmath> 12 #include<queue> 13 #define clc(a,b) memset(a,b,sizeof(a)) 14 #include <bits/stdc++.h> 15 const int maxn = 20005; 16 const int inf=0x3f3f3f3f; 17 const double pi=acos(-1); 18 typedef long long LL; 19 using namespace std; 20 //const LL MOD = 1e9+7; 21 22 LL PowMod(LL a,LL b,LL MOD){ 23 LL ret=1; 24 while(b){ 25 if(b&1) ret=(ret*a)%MOD; 26 a=(a*a)%MOD; 27 b>>=1; 28 } 29 return ret; 30 } 31 LL fac[100005]; 32 LL Get_Fact(LL p){ 33 fac[0]=1; 34 for(int i=1;i<=p;i++) 35 fac[i]=(fac[i-1]*i)%p; 36 } 37 LL Lucas(LL n,LL m,LL p){ 38 LL ret=1; 39 while(n&&m){ 40 LL a=n%p,b=m%p; 41 if(a<b) return 0; 42 ret=(ret*fac[a]*PowMod(fac[b]*fac[a-b]%p,p-2,p))%p; 43 n/=p; 44 m/=p; 45 } 46 return ret; 47 } 48 int main(){ 49 int t; 50 scanf("%d",&t); 51 while(t--){ 52 LL n,m,p; 53 scanf("%I64d%I64d%I64d",&n,&m,&p); 54 Get_Fact(p); 55 printf("%I64d ",Lucas(n+m,m,p)); 56 } 57 return 0; 58 }