Codeforces Beta Round #17 D. Notepad
主要问题是怎么求$a^b \ mod \ c$。
可以用欧拉定理降幂。
$$a^b \ mod \ c = a ^ {b \% \varphi(c) } \; mod \ c (b < \varphi(c))$$
$$a^b \ mod \ c = a ^ {b \% \varphi(c) + \varphi(c)} \; mod \ c (b \geqslant \varphi(c))$$
1 #include<cstdio> 2 #include<cstring> 3 #include<iostream> 4 using namespace std; 5 char B[1000005], A[1000005]; 6 typedef long long ll; 7 ll phi(int x) { 8 ll res = x; 9 for (int i = 2; i * i <= x; ++i) 10 if (x % i == 0) { 11 res = res / i * (i - 1); 12 while (x % i == 0) x /= i; 13 } 14 if (x > 1) res = res / x * (x - 1); 15 return res; 16 } 17 ll pow(ll a, ll b, ll m) { 18 ll res = 1; 19 for (a %= m; b; b >>= 1, a = (a * a) % m) 20 if (b & 1) res = (res * a) % m; 21 return res; 22 } 23 int c; 24 ll a, a_1, b; 25 int main() { 26 scanf("%s", A); scanf("%s", B); scanf("%d", &c); 27 int lal = strlen(A), lbl = strlen(B); 28 for (int i = 0; i < lal; ++i) { 29 a = (a * 10 + A[i] - '0') % c; 30 } 31 for (int i = lal - 1; ~i; --i) { 32 if (A[i] == '0') A[i] = '9'; 33 else {--A[i]; break;} 34 } 35 for (int i = 0; i < lal; ++i) 36 a_1 = (a_1 * 10 + A[i] - '0') % c; 37 int flag = 0, p = phi(c); 38 for (int i = 0; i < lbl; ++i) { 39 b = (b * 10 + B[i] - '0'); 40 if (b >= p) flag = 1; 41 b %= p; 42 } 43 if (flag) b += p; 44 ll ans = pow(a, b - 1, c) * a_1 % c; 45 printf("%lld\n", ans ? ans : c); 46 return 0; 47 }
