UVA 11582 Colossal Fibonacci Numbers! 数学

n比较小,最多n*n就回出现循环节....

Colossal Fibonacci Numbers!

Time Limit: 1000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu

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Description

Problem F: 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 â‰¥ 0

Your task is to compute some values of this sequence.

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

For each test case, output a single line containing the remainder of f (a^b) upon division by n.

Sample input

31 1 22 3 100018446744073709551615 18446744073709551615 1000

Sample output

121250

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Zachary Friggstad

Source

Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Examples
Root :: Prominent Problemsetters :: Zachary Friggstad

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#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef unsigned long long int uLL;uLL d[2000200],dn;uLL A,B,M;uLL quickpow(){    uLL e=1;    A=A%dn;    while(B)    {        if(B%2)        {            e=(e*A)%dn;        }        A=(A*A)%dn;        B/=(uLL)2;    }    return e;}int main(){    int T_T;    cin>>T_T;    while(T_T--)    {        dn=0;        cin>>A>>B>>M;        if(M==1)        {            puts("0"); continue;        }        uLL a=(uLL)1,b=(uLL)1;        uLL la=a,lb=b;        d[dn++]=a;d[dn++]=b;        while(true)        {            a=(la+lb)%M;            b=(lb+a)%M;            if(d[0]==a&&d[1]==b) break;            d[dn++]=a;d[dn++]=b;            la=a; lb=b;        }        cout<<d[quickpow()-1]<<endl;    }    return 0;}