指數迴圈節(降冪)
阿新 • • 發佈:2018-12-17
指數迴圈節
在有些題目中我們需要對指數進行降冪處理才能計算。比如計算
其中和
這裡由於很大,所以需要進行降冪。那麼實際上有如下降冪公式
給定,和的值,求的值,其中,
程式碼:
#include <iostream> #include <string.h> #include <stdio.h> using namespace std; const int N = 1000005; typedef long long LL; char str[N]; int phi(int n)//求φ(n) { int rea = n; for (int i = 2; i * i <= n; i++) { if (n % i == 0) { rea = rea - rea / i; while (n % i == 0) { n /= i; } } } if (n > 1) { rea = rea - rea / n; } return rea; } LL multi(LL a, LL b, LL m)//快乘a*b%m { LL ans = 0; a %= m; while (b) { if (b & 1) { ans = (ans + a) % m; b--; } b >>= 1; a = (a + a) % m; } return ans; } LL quick_mod(LL a, LL b, LL m)//快速冪a^b%mod { LL ans = 1; a %= m; while (b) { if (b & 1) { ans = multi(ans, a, m); b--; } b >>= 1; a = multi(a, a, m); } return ans; } void Solve(LL a, char str[], LL c)//a^str%c { LL len = strlen(str); LL ans = 0; LL p = phi(c); if (len <= 15) { for (int i = 0; i < len; i++) { ans = ans * 10 + str[i] - '0'; } } else { for (int i = 0; i < len; i++) { ans = ans * 10 + str[i] - '0'; ans %= p; } ans += p; } printf("%I64d\n", quick_mod(a, ans, c)); } int main() { LL a, c; while (~scanf("%I64d%s%I64d", &a, str, &c)) { Solve(a, str, c); } return 0; }