ASC7 Problem E. Strange Limit

题目大意

求$p^{p^{p^{.^{.^{.}}}}} \mod m!$

简要题解

由欧拉定理，有对任意$a,b,c$当$b$足够大时，有$a^b \equiv a^{b \mod \varphi(c) +\varphi(c)} \mod c$，设$x$为对$c=m!$取模的所求，$y$为对$c=\varphi(m!)$的所求，可以得到$x=a^{y+\varphi(c)}\mod c$，递归做即可。

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 namespace my_header {
  4 #define pb push_back
  5 #define mp make_pair
  6 #define pir pair<int, int>
  7 #define vec vector<int>
  8 #define pc putchar
  9 #define clr(t) memset(t, 0, sizeof t)
 10 #define pse(t, v) memset(t, v, sizeof t)
 11 #define bl puts("")
 12 #define wn(x) wr(x), bl
 13 #define ws(x) wr(x), pc(' ')
 14     const int INF = 0x3f3f3f3f;
 15     typedef long long LL;
 16     typedef double DB;
 17     inline char gchar() {
 18         char ret = getchar();
 19         for(; (ret == '\n' || ret == '\r' || ret == ' ') && ret != EOF; ret = getchar());
 20         return ret; }
 21     template<class T> inline void fr(T &ret, char c = ' ', int flg = 1) {
 22         for(c = getchar(); (c < '0' || '9' < c) && c != '-'; c = getchar());
 23         if (c == '-') { flg = -1; c = getchar(); }
 24         for(ret = 0; '0' <= c && c <= '9'; c = getchar())
 25             ret = ret * 10 + c - '0';
 26         ret = ret * flg; }
 27     inline int fr() { int t; fr(t); return t; }
 28     template<class T> inline void fr(T&a, T&b) { fr(a), fr(b); }
 29     template<class T> inline void fr(T&a, T&b, T&c) { fr(a), fr(b), fr(c); }
 30     template<class T> inline char wr(T a, int b = 10, bool p = 1) {
 31         return a < 0 ? pc('-'), wr(-a, b, 0) : (a == 0 ? (p ? pc('0') : p) : 
 32             (wr(a/b, b, 0), pc('0' + a % b)));
 33     }
 34     template<class T> inline void wt(T a) { wn(a); }
 35     template<class T> inline void wt(T a, T b) { ws(a), wn(b); }
 36     template<class T> inline void wt(T a, T b, T c) { ws(a), ws(b), wn(c); }
 37     template<class T> inline void wt(T a, T b, T c, T d) { ws(a), ws(b), ws(c), wn(d); }
 38     template<class T> inline T gcd(T a, T b) {
 39         return b == 0 ? a : gcd(b, a % b); }
 40     template<class T> inline T fpw(T b, T i, T _m, T r = 1) {
 41         for(; i; i >>= 1, b = b * b % _m)
 42             if(i & 1) r = r * b % _m;
 43         return r; }
 44 };
 45 using namespace my_header;
 46 
 47 const int PRIME = 300000;
 48 int vis[PRIME], pcnt, pri[PRIME];
 49 
 50 void sieve() {
 51     for (int i = 2; i < PRIME; ++i)
 52         if (pri[i] == 0) {
 53             for (int j = i; j < PRIME; j += i)
 54                 pri[j] = 1;
 55             pri[++pcnt] = i;
 56         }
 57 }
 58 
 59 int gcd(int a, int b) {
 60     return b == 0 ? a : gcd(b, a % b);
 61 }
 62 
 63 LL getPhi(LL t) {
 64     LL ret = 1;
 65     for (int i = 1; i <= pcnt; ++i) {
 66         if (t % pri[i] == 0) {
 67             ret *= pri[i] - 1;
 68             t /= pri[i];
 69             while (t % pri[i] == 0) {
 70                 t /= pri[i];
 71                 ret *= pri[i];
 72             }
 73         }
 74     }
 75     return t == 1 ? ret : t - 1;
 76 }
 77 
 78 LL solve(LL a, LL b) {
 79     if (b == 1)
 80         return 0;
 81     int phi = getPhi(b);
 82     //wt(phi);
 83     return fpw(a, solve(a, phi) + phi, b);
 84 }
 85 
 86 int main() {
 87 #ifdef lol
 88     freopen("E.in", "r", stdin);
 89     freopen("E.out", "w", stdout);
 90 #else
 91     freopen("limit.in", "r", stdin);
 92     freopen("limit.out", "w", stdout);
 93 #endif
 94     sieve();
 95     int p, m;
 96     fr(p, m);
 97     LL t = 1;
 98     for (int i = 1; i <= m; ++i)
 99         t = t * i;
100     wt(solve(p, t));
101 
102     return 0;
103 }

 

posted @ 2017-02-16 16:53  ichneumon  阅读(287)  评论(0)  编辑  收藏  举报