1. 前置知识
扩展欧几里得：对于不完全为\(0\)的非负整数\(a,b\)，\(\gcd(a,b)\)表示\(a,b\)的最大公约数，必然存在整数对\(x,y\)使得\(\gcd(a,b)=ax+by\)。
费马小定理:\(a^{p-1}\equiv 1\pmod p\)
逆元：
求解公式\((a/b)\% p\)时，因\(b\)可能会过大，会出现爆精度的情况，所以需变除法为乘法：
设\(c\)是\(b\)的逆元，则有\(b*c≡1\pmod p\)；
则\((a/b)\%p\equiv (a/b)*1\%p\equiv (a/b)*b*c\%p\equiv a*c \pmod p\);
即\(a/b\equiv a\times c\pmod p\)。
公式：
\(b*c\equiv 1\pmod p\)
\(b\equiv c^{-1} \pmod p\)
费马小定理求逆元：因为\(x^{p-1}\equiv 1\pmod p\)，提取一个\(x\)得\(x^{p-1}\equiv x\times x^{p-2}\equiv 1\pmod p\)，即\(x\)的逆元为\(x^{p-2}\)。注意：根据费马小定理，\(x\)与\(p\)互质时才可以用此方法。
扩展欧几里得求逆元：从\(b\times c\equiv 1\pmod m\)可以得出关系式：\(k\times m+1=b\times c\Rightarrow b\times c+(-k)\times m=1\)完全符合扩展欧几里得算法。
线性递推求解逆元：
inv[1]初始化为1
递推公式为：inv[i]=(p-p/i)*inv[p%i]%p
//其中p为模数，i为要求逆元的数，inv[i]为i的逆元
证明：
设\(t=m/i,k=m\%i\);
\(\Rightarrow t\times i+k\equiv 0\pmod m\)
\(\Rightarrow -t\times i-k\equiv 0\pmod m\)
\(\Rightarrow -t\times i\equiv k\pmod m\)
\(\Rightarrow -t\times 1/k\equiv 1/i\pmod m\)
\(\because 1/i\times i\equiv 1\pmod m\)
\(\therefore\)可以用\(inv[i]\)代替\(1/i\)
\(\Rightarrow -t*inv[k]\equiv inv[i]\pmod m\)
再替换\(t,k\)得
\(\Rightarrow inv[i]=(m-m/i)\times inv[m\%i]%m\)
公式得证。
二项式定理：
\((a+b)^n=\sum_{k=0}^{n}C_{n}^{k}a^{n-k}b^{k}\)
证明：
假设当\(n=m\)的时候二项式定理成立，那么当\(n=m+1\)时：
\[(a+b)^{m+1}\\ =a(a+b)^{m}+b(a+b)^{m}\\ =a\sum_{k=0}^{m} C_{m}^{k}a^{m-k}b^{k}+b\sum_{k=0}^{m} C_{m}^{k}a^{m-k}b^{k}\\ =a(\sum_{k=1}^{m} C_{m}^{k}a^{m-k}+a^{m})+b(\sum_{k=1}^{m-1} C_{m}^{k}a^{m-k}b^{k}+b^{m})\\ =a^{m+1}+b^{m+1}+\sum_{k=1}^{m}C_{m}^{k}a^{m-k+1}+\sum_{k=1}^{m}C_{m}^{k-1}a^{m-k+1}b^{k}\\ =a^{m+1}+b^{m+1}+\sum_{k=1}^{m}[(C_{m}^{k}+C_{m}^{k-1})]a^{m-k+1}b^{k}\\ =a^{m+1}+b^{m+1}+\sum_{k=1}^{m}C_{m+1}^{k}a^{m-k+1}b^{k}\\ =\sum_{k=0}^{0}C_{m+1}^{k}a^{m-k+1}b^{k}+\sum_{k=m+1}^{m+1}C_{m+1}^{k}a^{m-k+1}b^{k}+\sum_{k=1}^{m}C_{m+1}^{k}a^{m-k+1}b^{k}\\ =\sum_{k=0}^{m+1}C_{m+1}^{k}a^{m+1-k}b^{k}\\ =\sum_{k=0}^{m}C_{n}^{k}a^{n-k}b^{k} \]2. Lucas定理
概述：
对于非负整数\(m,n\)和质数\(p\)，组合数\(C_{n}^{m}\equiv \prod_{i=0}^{k}C_{n_{i}}^{m_{i}}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)式1
其中\(n\)的\(k\)位\(p\)进制展开式:\(n=n_{k}p^{k}+...+n_{1}p^{1}+n_{0}p^{0}\)
其中\(m\)的\(k\)位\(p\)进制展开式:\(m=m_{k}p^{k}+...+m_{1}p^{1}+m_{0}p^{0}\)
式1成立，则：\(C_{n}^{m}=C_{n_{k}}^{m_{k}}\times ...\times C_{n_{1}}^{m_{1}}\times C_{n_{0}}^{m_{0}}=\prod_{i=1}^{k}C_{n_{i}}^{m_{i}}\times C_{n_{0}}^{m_{0}}\)
其中\(\prod_{i=1}^{k}C_{n_{i}}^{m_{i}}=C_{\left \lfloor \frac{n}{p} \right \rfloor}^{\left \lfloor \frac{m}{p} \right \rfloor}\,\,\,,C_{n_{0}}^{m_{0}}=C_{n \mod p}^{m \mod p} \,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)
所以可得\(C_{n}^{m}=C_{\left \lfloor \frac{n}{p} \right \rfloor}^{\left \lfloor \frac{m}{p} \right \rfloor}\times C_{n \mod p}^{m \mod p}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\)式2
当\(n<m\)时，\(C_{n}^{m}=0\)。
证明：
设\(x\)是任意小于\(p\)的正整数，则：\(C_{p}^{x}=\frac{p!}{x!(p-x)!}=\frac{p\cdot (p-1)!}{x\cdot (x-1)!\cdot (p-x)!}=\frac{p}{x}\cdot C_{p-1}^{x-1}\)
由于\(x<p\)且\(p\)为质数，所以\(x\)存在模\(p\)意义下的逆元，故：\(C_{p}^{x}\equiv p\cdot inv(x)\cdot C_{p-1}^{x-1}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)
显然等号右边的式子是\(p\)的倍数，故\(C_{p}^{x}\equiv 0\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)
以上证明当\(x<p\)时，\(C_{p}^{x}=0\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)。
已知二项式定理：\((1+x)^{n}=\sum_{i=0}^{n}C_{n}^{i}x^{i}\)。由于\((1+x)^{p}=\sum_{i=0}^{p}C_{p}^{i}\)对于\(i\in [1,p-1]\)的项模\(p\)都为\(0\)，所以\((1+x)^{p}\equiv C_{n}^{n}x^{p}+c_{n}^{0}x^{0}\equiv x^{p}+1\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)。
设\(\begin{cases} \left \lfloor \frac{m}{p} \right \rfloor=q_{m}\\\left \lfloor \frac{n}{p} \right \rfloor=q_{n} \end{cases}\),\(\begin{cases} m\mod p=r_{m}\\n\mod p=r_{n} \end{cases}\)，则\(\begin{cases} m=q_{m}\cdot p+r_{m}\\n=q_{n}\cdot p+r_{n} \end{cases}\)。
则：
\[(1+x)^{n}\\ =(1+x)^{q_{n}\cdot p+r_{n}}\\ =[(1+x)^{p}]^{q_{n}}\cdot (1+x)^{r_{n}}\equiv (1+x^{p})^{q_{n}}\cdot (1+x)^{r_{n}}\\ =\sum_{i=0}^{q_{n}}C_{q_{n}}^{i}x^{i\cdot p}\sum_{j=0}^{r_{n}}C_{r_{n}}^{j}x^{j}\\ =\sum_{i=0}^{q_{n}}\sum_{j=0}^{r_{n}}C_{q_{n}}^{i}C_{r_{n}}^{j}x^{i\cdot p+j}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p \]则设\(k=i\cdot p+j\),\(k\in[0,n]\)，得\((1+x)^{n}\equiv \sum_{k=0}^{n}C_{q_{n}}^{\left \lfloor \frac{k}{p} \right \rfloor}C_{r_{n}}^{k\mod p}x^{k}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)
由二项式定理，得\((1+x)^{n}\equiv \sum_{k=0}^{n}C_{n}^{k}x^{k} \equiv \sum_{k=0}^{n}C_{q_{n}}^{\left \lfloor \frac{k}{p} \right \rfloor}C_{r_{n}}^{k\mod p}x^{k}\,\,\,\,\,\,\,\,\,\,\,\,\pmod p\)
令\(k=m\)、等式两边消除\(x^{k}\)单取二项式系数得\(C_{n}^{m}\equiv C_{q_{n}}^{\left \lfloor \frac{m}{p} \right \rfloor}C_{r_{n}}^{m\mod p}=C_{\left \lfloor \frac{n}{p} \right \rfloor}^{\left \lfloor \frac{m}{p} \right \rfloor}C_{n\mod p}^{m\mod p}\)
定理得证。
3.总结
（自我感觉）挺难的一个算法。
参考资料：
Shaoxing No.1 Middle School 内部学习资料
https://www.cnblogs.com/Lrefrain/p/11353765.html
https://blog.csdn.net/jk_chen_acmer/article/details/79305269
https://blog.csdn.net/jk_chen_acmer/article/details/79297068
https://blog.csdn.net/jk_chen_acmer/article/details/79271069
模板题：P3807
套公式即可。
#include<bits/stdc++.h>
using namespace std;
long long t;
int ksm(long long a,long long b,long long p)
{
long long ans=1,x=a;
while(b)
{
if(b&1) ans*=x,ans%=p;
x*=x;x%=p;b>>=1;
}
return ans%p;
}
long long C(long long m,long long n,long long p)
{
if(n<m) return 0;
if(m>n-m) m=n-m;
long long a=1,b=1;
for(long long i=0;i<m;++i) {a=a%p*(n-i)%p;b=b%p*(i+1)%p;}
return a%p*ksm(b,p-2,p)%p;
}
long long lucas(long long n,long long m,long long p)
{
if(m==0) return 1;
return C(n/p,m/p,p)%p*C(n%p,m%p,p)%p;
}
void work()
{
long long n,m,p;
cin>>n>>m>>p;
cout<<lucas(n,n+m,p)%p<<"\n";
}
int main()
{
cin>>t;
while(t--) work();
return 0;
}