UVA - 11582 Colossal Fibonacci Numbers! （巨大的斐波那契数！）

题意：输入两个非负整数a、b和正整数n(0<=a,b<2⁶⁴,1<=n<=1000)，你的任务是计算f(a^b)除以n的余数，f(0) = 0, f(1) = 1,且对于所有非负整数i，f(i + 2) = f(i + 1) + f(i)。

分析：

1、对于某个n取余的斐波那契序列总是有周期的，求出每个取值的n下的斐波那契序列和周期。

2、a^b对T[n]取余，即可确定对n取余的斐波那契序列中f(a^b)的位置。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b) {
    if(fabs(a - b) < eps)  return 0;
    return a < b ? -1 : 1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
vector<int> v[MAXN];
int T[MAXN];//周期
void init(){
    for(int i = 2; i <= 1000; ++i){
        v[i].push_back(0);
        v[i].push_back(1);
        for(int j = 2; ; ++j){
            v[i].push_back((v[i][j - 1] + v[i][j - 2]) % i);
            if(v[i][j] == 1 && v[i][j - 1] == 0){
                T[i] = j - 1;
                break;
            }
        }
    }
}
ULL Q_POW(ULL a, ULL b, int n){
    ULL ans = 1ULL;
    ULL tmp = a;
    while(b){
        if(b & 1){
            ans = (ans * tmp) % n;
        }
        tmp = (tmp * tmp) % n;
        b >>= 1;
    }
    return ans;
}
int main(){
    int N;
    scanf("%d", &N);
    init();
    while(N--){
        ULL a, b, n;
        scanf("%llu%llu%llu", &a, &b, &n);
        if(a == 0 || n == 1){
            printf("0\n");
            continue;
        }
        ULL ans = Q_POW(a % T[n], b, T[n]);
        printf("%d\n", v[n][ans]);
    }
    return 0;
}