Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).
Input
There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.
Output
For each testcase, output an integer, denotes the result of A^B mod C.
Sample Input
3 2 4 2 10 1000
Sample Output
1 24
思路:欧拉降幂板子
typedef long long LL; LL A, C; char B[10000005]; LL getEuler(LL x) { LL ans = x; for(LL i = 2; i*i <= x; ++i) { if(x % i == 0) { ans = ans / i * (i-1); while(x % i == 0) x /= i; } } if(x > 1) ans = ans / x * (x-1); return ans; } LL quickPow(LL a, LL b, LL p) { // a^b (modp) LL ret = 1; while(b) { if(b & 1) ret = (ret * a) % p; a = (a * a) % p; b >>= 1; } return ret; } int main() { ios::sync_with_stdio(false), cin.tie(NULL); while(cin >> A >> B >> C) { LL phi = getEuler(C); LL num = 0; int len = strlen(B); for(int i = 0; i < len; ++i) { num = (num * 10 + B[i] - '0'); if(num >= phi) break; } if(num >= phi) { num = 0; for(int i = 0; i < len; ++i) num = (num * 10 + B[i] - '0') % phi; cout << quickPow(A, num+phi, C) << " "; } else { cout << quickPow(A, num, C) << " "; } } return 0; }