题目链接 : http://acm.hdu.edu.cn/showproblem.php?pid=4828

 

Catalan数的公式为 C[n+1] = C[n] * (4 * n + 2) / (n + 2)

题目要求对M = 1e9+7 取模

利用乘法逆元将原式中除以(n+2)取模变为对(n+2)逆元的乘法取模

C[n+1] = C[n] * (4 * n + 2) * Pow(n+2, MOD-2) % MOD

其中Pow用快速幂解决

#include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long LL;const int MAXN = 1e6+10;const int MOD = 1e9+7;LL C[MAXN];LL QuickPow(LL x, LL n){    LL ans = 1;    while(n) {        if(n & 1) ans = (ans * x) % MOD;        x = (x * x) % MOD;        n /= 2;    }    return ans;}void Pre(){    C[1] =  1;    for(int i = 2; i <= MAXN; i++) {        C[i] = C[i-1] * (4 * i - 2) % MOD * QuickPow(i + 1, MOD-2) % MOD;    }}int main(){    Pre();    int t;    int n;    scanf("%d", &t);    for(int cas = 1; cas <= t; cas++) {        scanf("%d", &n);        printf("Case #%d:\n%I64d\n", cas, C[n]);    }    return 0;}