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<2,1<=n<=1000)，你的任务是计算f(a)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对T[n]取余，即可确定对n取余的斐波那契序列中f(a)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;

}