BZOJ1951 [Sdoi2010]古代猪文

这道题是各种数论搞在一起的题目。。。

首先由burnside引理可以知道答案是ans = (G^sigma(C(n, d))) % MOD

然后由费马小定理，ans =  (G^（sigma(C(n, d)) % (MOD - 1)）) % MOD

之后把MOD - 1分解为2 * 3 * 4679 * 35617，用Lucas定理分别求出sigma(C(n, d)) % m，其中m = 2, 3, 4679, 35617

最后再用中国剩余定理合起来快速幂一下就好辣！

 

 1 /**************************************************************
 2     Problem: 1951
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:76 ms
 7     Memory:2772 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12  
13 using namespace std;
14 typedef long long LL;
15 const LL p = 999911659;
16 const int x[4] = {2, 3, 4679, 35617};
17  
18 LL mod[4][50000], a[50000], b[5], cnt, n, g, y, z;
19  
20 inline LL pow(LL a, LL b, LL p){
21     LL res = 1;
22     while (b){
23         if (b & 1) res *= a, res %= p;
24         a *= a, a %= p;
25         b >>= 1;
26     }
27     return res;
28 }
29  
30 LL extend_gcd(int a, int b, LL &x, LL &y){
31     if (!b){
32         x = 1, y = 0;
33         return a;
34     }
35     LL res = extend_gcd(b, a % b, y, x);
36     y -= a / b * x;
37     return res;
38 }
39  
40 inline LL GCD(int a, int p){
41     LL x, y, d = extend_gcd(a, p, x, y);
42     while (x < 0) x += p;
43     return x;
44 }
45  
46 inline LL C(int i, int n, int k, int p){
47     return mod[i][n] * GCD(mod[i][n - k] * mod[i][k] % p, p) % p;
48 }
49  
50 inline LL Lucas(int i, int n, int k, int p){
51     LL res = 1;
52     while (n && k){
53         res *= C(i, n % p, k % p, p), res %= p;
54         if (!res) return 0;
55         n /= p, k /= p;
56     }
57     return res;
58 }
59  
60 inline LL Chinese(){
61     LL M = 1, res = 0;
62     for (int i = 0; i < 4; ++i) M *= x[i];
63     for (int i = 0; i < 4; ++i){
64         LL w = M / x[i], x1, y1, d = extend_gcd(w, x[i], x1, y1);
65         res += x1 * w * b[i], res %= M;
66     }
67     return (res + M) % M;
68 }
69  
70 int main(){
71     scanf("%lld%lld", &n, &g);
72     g %= p;
73     if (!g){
74         printf("0\n");
75         return 0;
76     }
77     for (int i = 0; i < 4; ++i){
78         mod[i][0] = 1;
79         for (int j = 1; j <= x[i]; ++j)
80             mod[i][j] = (mod[i][j - 1] * j) % x[i];
81     }
82     int maxi = (int) sqrt(n);
83     for (int i = 1; i <= maxi; ++i)
84         if (!(n % i)){
85             a[++cnt] = i;
86             if (i * i != n) a[++cnt] = n / i;
87         }
88          
89     for (int i = 1; i <= cnt; ++i)
90         for (int j = 0; j < 4; ++j){
91             b[j] += Lucas(j, n, a[i], x[j]);
92             if (b[j] > x[j]) b[j] -= x[j];
93         }
94     int z = Chinese();
95     printf("%lld\n", pow(g, z, p));
96     return 0;
97 }
View Code

 

posted on 2014-09-28 17:25  Xs酱~  阅读(282)  评论(1)  编辑  收藏  举报