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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
    #endif
    
    typedef 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
    ", (int)ans);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/moonbay/p/4139718.html
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