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bjfu1238 卡特兰数取余

题目就是指定n,求卡特兰数Ca(n)%m。求卡特兰数有递推公式、通项公式和近似公式三种,因为要取余,所以近似公式直接无法使用,递推公式我简单试了一下,TLE。所以只能从通项公式入手。

Ca(n) = (2*n)! / n! / (n+1)!

思想就是把Ca(n)质因数分解,然后用快速幂取余算最后的答案。不过,算n!时如果从1到n依次质因数分解,肯定是要超时的,好在阶乘取余有规律,不断除素因子即可。

最后还是擦边过,可能筛法写得一般吧,也算是题目要求太柯刻。

/* * Author    : ben */#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>#include <stack>#include <string>#include <vector>#include <deque>#include <list>#include <functional>#include <numeric>#include <cctype>using namespace std;#ifdef ON_LOCAL_DEBUG#else#endiftypedef long long LL;const int MAXN = 80000;const int N = 1000001;bool isPrime[N + 3];//多用两个元素以免判断边界int pn, pt[MAXN], num[MAXN];void init_prime_table() {    memset(isPrime, true, sizeof(isPrime));    int p = 2, q, del;    double temp;    while (p <= N) {        while (!isPrime[p]) {        p++;        }        if (p > N) {//已经结束            break;        }        temp = (double) p;        temp *= p;        if (temp > N)            break;        while (temp <= N) {            del = (int) temp;            isPrime[del] = false;            temp *= p;        }        q = p + 1;        while (q < N) {            while (!isPrime[q]) {    q++;    }            if (q >= N) { break;}            temp = (double) p;            temp *= q;            if (temp > N)    break;            while (temp <= N) {                del = (int) temp;                isPrime[del] = false;                temp *= p;            }            q++;        }        p++;    }    pn = 0;    for (int i = 2; i <= N; i++) {        if (isPrime[i]) {            pt[pn++] = i;        }    }}int modular_exp(int a, int b, int c) {    LL res, temp;    res = 1 % c, temp = a % c;    while (b) {        if (b & 1) {            res = res * temp % c;        }        temp = temp * temp % c;        b >>= 1;    }    return (int) res;}void getNums(int n) {    int x = 2 * n;    int len = pn;    for (int i = 0; i < len && pt[i] <= x; i++) {        int r = 0, a = x;        while (a) {            r += a / pt[i];            a /= pt[i];        }        num[i] = r;    }    x = n;    for (int i = 0; i < len && pt[i] <= x; i++) {        int r = 0, a = x;        while (a) {            r += a / pt[i];            a /= pt[i];        }        num[i] -= r;    }    x = n + 1;    for (int i = 0; i < len && pt[i] <= x; i++) {        int r = 0, a = x;        while (a) {            r += a / pt[i];            a /= pt[i];        }        num[i] -= r;    }}int main() {#ifdef ON_LOCAL_DEBUG    freopen("data.in", "r", stdin);//    freopen("test.in", "r", stdin);//    freopen("data.out", "w", stdout);#endif    int n, m;    init_prime_table();    while (scanf("%d%d", &n, &m) == 2) {        getNums(n);        LL ans = 1;        int n2 = 2 * n;        for (int i = 0; ans && (i < pn) && pt[i] <= n2; i++) {            if (num[i] > 0) {                ans = ans * (LL) modular_exp(pt[i], num[i], m);                ans %= m;            }        }        printf("%d\n", (int)ans);    }    return 0;}

 

bjfu1238 卡特兰数取余