学步园

预处理求出n分别为1-1000时斐波那契模数列的周期，之后求出a^b是周期中第几个即可。

复习了模算术、幂取模，顺便学习了分治法取模的迭代版本。

Run Time: 0.072s

#define UVa  "LT10-1.11582.cpp"		//Colossal Fibonacci Numbers!
char fileIn[30] = UVa, fileOut[30] = UVa;

#include<cstring>
#include<cstdio>
#include<algorithm>

using namespace std;


//Global Variables. Reset upon Each Case!
typedef unsigned long long ULL;
const int maxn = 1000 + 5;
int m[maxn];
unsigned long long f[maxn][3100];
unsigned long long a,b;
int n;
/////


int power(ULL a, ULL b, int t) {
    ULL result = 1;
    while(b) {
        if(b%2) {
            result = (a*result)%t;
        }
        a = (a*a)%t;
        b/=2;
    }
    return result;
}


int main() {
    for(n = 2; n <= 1000; n ++) {
        int tmp = 0;
        f[n][0] = 0, f[n][1] = 1;
        for(int i = 2; ; i ++) {
            f[n][i] = (f[n][i-2]%n + f[n][i-1]%n) % n;
            if(f[n][i] == 1 && f[n][i-1] == 0) {
                m[n] = i-1;
                break;
            }
        }
    }

    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%llu%llu%d", &a, &b, &n);
        if(a == 0 || n == 1) printf("0\n");
        else printf("%d\n", f[n][power(a%m[n], b, m[n])]);
    }

    return 0;
}