Uva 11582 Colossal Fibonacci Numbers! (循环节、幂取模、打表)

The i’th Fibonacci number f(i) is recursively defined in the following way:
• f(0)=0$f(0)=0$ and f(1)=1$f(1)=1$
•f(i+2)=f(i+1)+f(i)$f(i+2)=f(i+1)+f(i)$for every i≥0$i\ge 0$

Your task is to compute some values of this sequence.
Input
Input begins with an integer t≤10,000$t\le 10,000$, the number of test cases.
Each test case consists of three integers a, b, n where 0≤a,b<264$0\le a,b<2^{64}$
(a and b will not both be zero) and 1≤n≤1000$1\le n\le 1000$.
Output
For each test case, output a single line containing the remainder of f(ab$a^b$) upon division by n
Sample Input
3
1 1 2
2 3 1000
18446744073709551615 18446744073709551615 1000
Sample Output
1
21
250
ps: 这题数据非常大，必须通过找循环节来解决，f(i+2)=f(i+1)+f(i)$f(i+2)=f(i+1)+f(i)$,当f(i), f(i+1)与前面的数相同时，就会开始产生循环节。因为余数最多n种，所以最多n2$n^2$项就会出现重复。鉴于数据规模实在过于庞大(106∗103=109${10}^6\ast {10}^3={10}^9$),单纯靠每次找循环节和快速幂会造成TLE(我上交tle了(,,•́ . •̀,,) ), 所以还需要将1~1000取模的所有情况打表，就是记录他们的循环节。
TLE code:

#include<iostream>
using namespace std;
const int maxn = 1e6+10;
int fib[maxn];
int pow(unsigned long long a, unsigned long long b, int c)
{//幂取模 a^b % c
    int ans = 1;
    a = a % c;
    while(b > 0)
    {
        if(b & 1) ans = ((a % c) * (ans % c)) % c;
        a = (a * a) % c;
        b = b >> 1;
    }
    return ans;
}
int main()
{
    //虽然双加速但也超时
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    unsigned long long a, b;
    int c;
    while(n--)
    {
        cin>>a>>b>>c;
        int maxt = c * c + 4;
        fib[0] = 0 , fib[1] = 1;
        for(int i = 2; i < maxt; i++)fib[i] = (fib[i - 1] % c + fib[i - 2] % c) % c;
        int temp = 1;
        for(int i = 2; i < maxt; i++){
            if(fib[0] == fib[i] && fib[1] == fib[i+1])
            {
                temp = i;
                break;
            }
        }
        int tt = pow(a, b, temp);
        if(a == 0 || c == 1)cout<<0<<endl;
        else cout<<fib[tt % temp]<<endl;
    }
    return 0;
}

改好后:

#include<iostream>
#include<vector>
using namespace std;
vector<int> fib[1004];//保存循环节
void init()
{//打表保存循环节
    for(int i = 2; i <= 1000; i++)
    {
        int mod = i;
        int a = 0, b = 1;
        int c = (a + b) % mod;
        fib[i].push_back(a);
        fib[i].push_back(b);
        fib[i].push_back(c);
        while(!(b == 0 && c == 1))
        {
            a = b;
            b = c;
            c = ((a % mod) + (b % mod)) % mod;
            fib[i].push_back(c);
        }
        fib[i].pop_back();
        fib[i].pop_back();
    }
}
int pow(unsigned long long a, unsigned long long b, int n)
{//幂取模
    int ans = 1;
    while(b > 0)
    {
        if(b & 1)ans = ((a % n) * (ans % n)) % n;
        a = ((a % n) * (a % n)) % n;
        b = b >> 1;
    }
    return ans;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin>>T;
    init();
    while(T--)
    {
        unsigned long long a, b;
        int n;
        cin>>a>>b>>n;
        if(a == 0 || n == 0){//f(0) = 0
            cout<<0<<endl;
            continue;
        }
        int t = pow(a, b, fib[n].size());
        cout<<fib[n][t]<<endl;
    }
    return 0;
}