Luogu3904 三只小猪 （组合数学，第二类斯特林数，高精）

即使\(n<=50\)，斯特林数也会爆long long。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define R(a,b,c) for(register int  a = (b); a <= (c); ++ a)
#define nR(a,b,c) for(register int  a = (b); a >= (c); -- a)
#define Max(a,b) ((a) > (b) ? (a) : (b))
#define Min(a,b) ((a) < (b) ? (a) : (b))
#define Fill(a,b) memset(a, b, sizeof(a))
#define Abs(a) ((a) < 0 ? -(a) : (a))
#define Swap(a,b) a^=b^=a^=b
#define ll long long

#define ON_DEBUG

#ifdef ON_DEBUG

#define D_e_Line printf("\n\n----------\n\n")
#define D_e(x)  cout << #x << " = " << x << endl
#define Pause() system("pause")
#define FileOpen() freopen("in.txt","r",stdin);

#else

#define D_e_Line ;
#define D_e(x)  ;
#define Pause() ;
#define FileOpen() ;

#endif

struct ios{
    template<typename ATP>ios& operator >> (ATP &x){
        x = 0; int f = 1; char c;
        for(c = getchar(); c < '0' || c > '9'; c = getchar()) if(c == '-')  f = -1;
        while(c >= '0' && c <= '9') x = x * 10 + (c ^ '0'), c = getchar();
        x*= f;
        return *this;
    }
}io;
using namespace std;

const int N = 1007;

int a[N], b[N];
string add(string A, string B){
	string S;
	Fill(a, 0), Fill(b, 0);
	int lenA = A.size(), lenB = B.size();
	R(i,0,lenA - 1) a[i] = A[lenA - i - 1] ^ '0';
	R(i,0,lenB - 1) b[i] = B[lenB - i - 1] ^ '0';
	int len = Max(lenA, lenB);
	R(i,0,len - 1){
		a[i] += b[i];
		a[i + 1] += a[i] / 10;
		a[i] %= 10;
	}
	if(a[len]) ++len;
	nR(i,len - 1,0) S += a[i] + '0';
	return S;
}
string mul(string A, int B){
	string S;
	int len = A.size();
	Fill(a, 0);
	R(i,0,len - 1) a[i] = A[len - i - 1] ^ '0';
	int res = 0;
	R(i,0,len - 1){
		a[i] = a[i] * B + res;
		res = a[i] / 10;
		a[i] = a[i] % 10;
	}
	while(res){
		a[len++] = res % 10;
		res /= 10;
	}
	nR(i,len - 1, 0) S += a[i] + '0';
	return S;
}
string f[107][107];
int n, m;
int main(){
	//FileOpen();
	int n, m;
	io >> n >> m;
	if(n < m){
		printf("0");
		return 0;
	}
	if(n == m){
		printf("1");
		return 0;
	}
    R(i,1,n){
    	f[i][1] = "1";
    	R(j,2,m){
    		f[i][j] = add(f[i - 1][j - 1], mul(f[i - 1][j], j));
    	}
    }

	cout << f[n][m];

    return 0;
}