hunnu11562:The Triangle Division of the Convex Polygon(第n个卡特兰数取模)

Problem description  A convex polygon with n edges can be divided into several triangles by some non-intersect diagonals. We denote d(n) is the number of the different ways to divide the convex polygon. For example,when n is 6,there are 14 different ways.Figure 1 shows such 14 ways. Then ,we get d(6)=14. 

Figure 1 when n=6, d(6)=14 .
Your task is that:for a given integer n, you’d tell us d(n). Because the d(n) may be a very large number, the result need to modulo m.
Input  There are multiple test cases. 
Each test case contains two postive integers (n,m) separated by a space in one line. 
The input will be terminated by the end of input file.
3<=n<=500,000 
1<=m<=45,000
Output  For each test case ,output an integer d(n)%m in one line.
Sample Input

6 103 10

Sample Output

41

Problem Source  2014哈尔滨理工大学秋季训练赛

题意：

一个正多边形分割成三角形有多少种不同的分法

思路：

不难看出是卡特兰数，实在是数学乏力，超时到身心疲惫，用分解质因子的方法需要980+ms，太不理想了，在别人那里看到了一个不错的代码，只要31ms，于是引用来做模板

地址：http://blog.csdn.net/kongming_acm/article/details/6361176

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define ls 2*i#define rs 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 1000005#define INF 0x3f3f3f3f#define EXP 1e-8#define rank rank1const int mod = 1000000007;int Pow(int x, int b){    int ret = 1;    for(int s = x; b; b >>= 1)    {        if(b & 1)            ret *= s;        s *= s;    }    return ret;}int PowMod(int x, int b, int p){    int ret = 1 % p;    for(int s = x % p; b; b >>= 1)    {        if(b & 1)        {            ret *= s;            ret %= p;        }        s *= s;        s %= p;    }    return ret;}pair<int, int> ExGcd(int a, int b){    if(a == 0) return make_pair(0, 1);    pair<int, int> s = ExGcd(b % a, a);    return make_pair(s.second - b / a * s.first, s.first);}int Inv(int a, int m) // ax == 1 mod m{// assert gcd(a, m) == 1    pair<int, int> s = ExGcd(a, m);// s.first * a + s.second * m == 1    return s.first < 0 ? s.first + m : s.first;}pair<int, int> FacMod(int n, int p, int k){// assert p > 1, k > 0    int pk = Pow(p, k);    int S[pk];    S[0] = 1 % pk;    for(int i = 1; i < pk; ++i)    {        S[i] = S[i - 1];        if(i % p != 0)        {            S[i] *= i;            S[i] %= pk;        }    }    int ret = 1 % pk, et = 0, ep = 0; // S[pk - 1]^et, p^ep    while(n)    {        et += n / pk;        ret *= S[n % pk];        ret %= pk;        ep += n / p;        n /= p;    }    ret *= PowMod(S[pk - 1], et, pk);    ret %= pk;    return make_pair(ret, ep);}int CRT(int x1, int m1, int x2, int m2) // Linear Congruence Equation{// assert gcd(m1, m2) == 1// let x == x1 * k1 + x2 * k2 mod m1*m2// k1 == 1 mod m1, k1 == 0 mod m2// k2 == 0 mod m1, k2 == 1 mod m2    pair<int, int> s = ExGcd(m1, m2);// s.first * m1 + s.second * m2 == 1    int k1 = s.second * m2, k2 = s.first * m1;    int t = (x1 * k1 + x2 * k2) % (m1 * m2);    return t < 0 ? t + m1 * m2 : t;}vector<pair<int, int> > Factorize(int n){    vector<pair<int, int> > ret;    for(int x = 2; x * x <= n; ++x)    {        if(n % x == 0)        {            int c = 0;            while(n % x == 0)            {                ++c;                n /= x;            }            ret.push_back(make_pair(x, c) );        }    }    if(n > 1) ret.push_back(make_pair(n, 1) );    return ret;}int Calc(int n, int p){    vector<pair<int, int> > factors = Factorize(p);    int x = 0, m = 1;    for(int i = 0; i < factors.size(); ++i)    {        // C(n, k) mod p[i]^k[i]        pair<int, int> f2n = FacMod(n * 2, factors[i].first, factors[i].second);        pair<int, int> fn = FacMod(n, factors[i].first, factors[i].second);        pair<int, int> fn1 = FacMod(n + 1, factors[i].first, factors[i].second);        int pk = Pow(factors[i].first, factors[i].second);        int t = f2n.first * Inv(fn.first, pk) % pk;        t = t * Inv(fn1.first, pk) % pk;        t = t * PowMod(factors[i].first, f2n.second - fn.second - fn1.second, pk) % pk;        x = CRT(x, m, t, pk);        m *= pk;    }    return x;}int main(){    int n, p;    while(scanf("%d%d", &n, &p) != -1)        printf("%d\n", Calc(n-2, p) );    return 0;}