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The Triangle Division of the Convex Polygon
题意:求 n 凸多边形可以有多少种方法分解成不相交的三角形,最后值模 m。
思路:卡特兰数的例子,只是模 m 让人头疼,因为 m 不一定是素数,所以不一定存在逆元。
解法:式子为f(n) = ( C( 2*(n-2), (n-2) ) / (n-1)) % m ;令 p = n-2, 式子可化为:f(p) = ((2*p)! / ( p! * (p+1)! ) ) % m;
对 s!分解质因素,统计个数。设小于等于 s 的素数为 p1, p2, p3, ... , pk;
则各个素因子个数为 :
for i = 1 to k q = s num(i) = 0 while q > 0 q = q / pi num(i) += q end while end for
所以,我们就可以统计出 f(p) 的素因子及个数,分子 + , 分母 - 。最后计算时用快速幂。
代码:
#include <climits> #include <cstdio> #include <cstring> #include <cctype> #include <cmath> #include <ctime> #include <cstdlib> #include <cstdarg> #include <iostream> #include <fstream> #include <iomanip> #include <sstream> #include <exception> #include <stdexcept> #include <memory> #include <locale> #include <bitset> #include <deque> #include <list> #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <algorithm> #include <iterator> #include <functional> #include <string> #include <complex> #include <valarray> using namespace std; typedef long long ll; const int N = 1e6+7; bool tag[N]; int p[N>>3]; int t; void prime() { t = 0; memset(tag, 0, sizeof tag); p[t++] = 2, tag[4] = 0; for(int i = 3; i < N; i += 2) { if(!tag[i]) p[t++] = i; for(int j = 0, k; j < t && (k = i * p[j]) < N; ++j) { tag[k] = 1; if(i % p[j] == 0) break; } } return ; } int n; ll m, ans; int zp[N>>3], mp[N>>3]; int tz, tp; int Factor(int q[], int u) { //分解 n! int i; for( i = 0; i < t && p[i] <= u; ++i) { int v = u; while(v) { v /= p[i]; q[i] += v; } } return i; } void cat(int n) { int nn = n + n; tz = tp = 0; memset(zp, 0, sizeof zp); memset(mp, 0, sizeof mp); tz = Factor(zp, nn); tp = Factor(mp, n); tp = Factor(mp, n+1); for(int i = 0; i < tp; ++i) zp[i] -= mp[i]; return ; } ll mult_mod(int a, int b, ll m) { ll res = 1LL, tt = (ll) a; while(b) { if(b&1) res = (res * tt) % m; tt = tt * tt % m; b >>= 1; } return res; } void solve() { n -= 2; cat(n); ans = 1LL; for(int i = 0; i < tz; ++i) { ans = (ans * mult_mod(p[i], zp[i], m)) % m; } printf("%I64d\n", ans); } int main() { #ifdef PIT freopen("c.in", "r", stdin); #endif // PIT prime(); while (~scanf("%d %I64d", &n, &m)) { solve(); } return 0; }
HOJ 13101 The Triangle Division of the Convex Polygon(数论求卡特兰数(模不为素数))
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原文地址:http://www.cnblogs.com/Duahanlang/p/4134631.html