UVA 11582 Colossal Fibonacci Numbers!(打表+快速幂）

Colossal Fibonacci Numbers!

The i'th Fibonacci number f (i) is recursively defined in the following way:

• f (0) = 0 and f (1) = 1
• f (i+2) = f (i+1) + f (i)  for every i ≥ 0

Your task is to compute some values of this sequence.

Input begins with an integer t ≤ 10,000, the number of test cases.Each test case consists of three integersa,b,n where 0 ≤ a,b < 2⁶⁴(a and b will not both be zero)and 1 ≤ n ≤ 1000.

For each test case, output a single linecontaining the remainder of f (a^b) upon division byn.

Sample input

31 1 22 3 100018446744073709551615 18446744073709551615 1000

Sample output

121250

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<limits.h>#include<vector>using namespace std;typedef unsigned long long ULL;//long long 挂ULL n,m,MOD;vector<int>f[1001];void init(){    for(int i=2;i<=1000;i++)    {        int mod=i;        int a=0,b=1,c=(a+b)%mod;        f[i].push_back(a);        f[i].push_back(b);        f[i].push_back(c);        while(!(b==0&&c==1))        {            a=b;            b=c;            c=(a%mod+b%mod)%mod;            f[i].push_back(c);        }        f[i].pop_back();        f[i].pop_back();    }}ULL quick_mod(ULL a,ULL b,ULL m)//快速幂缩小区间{     ULL ans = 1;     while(b)     {        if(b&1)        {            ans=((ans%m)*(a%m))%m;            b--;        }        b/=2;        a=((a%m)*(a%m))%m;     }     return ans;}int main(){    int t;    cin>>t;    init();//打表    while(t--)    {        cin>>n>>m>>MOD;        if(MOD==1)        {            cout<<0<<endl;            continue;        }        ULL mm=f[MOD].size();        ULL ans=quick_mod(n,m,mm);        cout<<f[MOD][ans]<<endl;    }    return 0;}