题意：问用不超过 m 颗种子放到 n 棵树中，有多少种方法。

析：题意可以转化为 x1 + x2 + .. + xn = m，有多少种解，然后运用组合的知识就能得到答案就是 C(n+m, m)。

然后就求这个值，直接求肯定不好求，所以我们可以运用Lucas定理，来分解这个组合数，也就是Lucas(n,m,p)=C(n%p,m%p)* Lucas(n/p,m/p,p)。

然后再根据费马小定理就能做了。

代码如下：

第一种：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10005;

const LL mod = 10000000000007;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};

const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

LL exgcd(LL a,LL b,LL &x,LL &y){LL d = a;if(b!=0){d=exgcd(b,a%b,y,x);y-=(a/b)*x;}else{x=1;y=0;}return d;}

LL mod_inverse(LL a,LL m){LL x,y;exgcd(a,m,x,y);return (m+x%m)%m; }

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

LL fact[100005];

LL mod_fact(LL n, LL p, LL &e){

    e = 0;

    if(!n)  return 1;

    LL res = mod_fact(n / p, p, e);

    e += n / p;

    if(n / p % 2 != 0)  return res * (p - fact[n%p]) % p;

    return res * fact[n%p] % p;

}

LL mod_comb(LL n, LL k, LL p){

    if(n < 0 || k < 0 || n < k) return 0;

    LL e1, e2, e3;

    LL a1 = mod_fact(n, p, e1);

    LL a2 = mod_fact(k, p, e2);

    LL a3 = mod_fact(n-k, p, e3);

    if(e1 > e2+e3)  return 0;

    return a1 * mod_inverse(a2*a3%p, p) % p;

}

int main(){

    fact[0] = 1;

    int T;  cin >> T;

    while(T--){

        LL p, m, n;

        scanf("%I64d %I64d %I64d", &n, &m, &p);

        for(int i = 1; i < p; ++i)  fact[i] = fact[i-1] * (LL)i % p;

        printf("%I64d\n", mod_comb(n+m, m, p));

    }

    return 0;

}

第二种：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10005;

const LL mod = 10000000000007;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};

const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

LL fact[100005];

LL p;

LL quick_pow(LL a, LL b){

    LL ans = 1LL;

    a %= p;

    while(b){

        if(b & 1)  ans = ans * a % p;

        a = a * a % p;

        b >>= 1;

    }

    return ans;

}

LL C(LL n, LL m){

    if(n < m)  return 0;

    return fact[n] * quick_pow(fact[m]*fact[n-m], p-2) % p;

}

LL lucas(LL n, LL m){

    if(!m)   return 1LL;

    return C(n%p, m%p) * lucas(n/p, m/p) % p;

}

int main(){

    fact[0] = 1;

    int T;  cin >> T;

    while(T--){

        LL m, n;

        scanf("%I64d %I64d %I64d", &n, &m, &p);

        for(int i = 1; i < p; ++i)  fact[i] = fact[i-1] * (LL)i % p;

        printf("%I64d\n", lucas(n+m, m));

    }

    return 0;

}

第三种：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <unordered_map>

//#include <tr1/unordered_map>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

//using namespace std :: tr1;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const LL LNF = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10005;

const LL mod = 10000000000007;

const int N = 1e6 + 5;

const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};

const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};

const int hr[]= {-2, -2, -1, -1, 1, 1, 2, 2};

const int hc[]= {-1, 1, -2, 2, -2, 2, -1, 1};

const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

LL p;

LL quick_pow(LL a, LL b){

    LL ans = 1LL;

    a %= p;

    while(b){

        if(b & 1)  ans = ans * a % p;

        a = a * a % p;

        b >>= 1;

    }

    return ans;

}

LL C(LL n, LL m){

    if(n < m)  return 0;

    LL a = 1, b = 1;

    while(m){

        a = a * n % p;

        b = b * m % p;

        --m;  --n;

    }

    return a * quick_pow(b, p-2) % p;

}

LL lucas(LL n, LL m){

    if(!m)   return 1LL;

    return C(n%p, m%p) * lucas(n/p, m/p) % p;

}

int main(){

    int T;  cin >> T;

    while(T--){

        LL m, n;

        scanf("%I64d %I64d %I64d", &n, &m, &p);

        printf("%I64d\n", lucas(n+m, m));

    }

    return 0;

}

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