HDU2813Interesting Fibonacci(斐波那契数列+循环节)

Interesting Fibonacci

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 694    Accepted Submission(s): 121


Problem Description

In mathematics, the Fibonacci numbers are a sequence of numbers named after Leonardo of Pisa, known as Fibonacci (a contraction of filius Bonaccio, "son of Bonaccio"). Fibonacci's 1202 book Liber Abaci introduced the sequence to Western European mathematics,
although the sequence had been previously described in Indian mathematics.

  The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc. In mathematical terms, it is defined by the following
recurrence relation:



That is, after two starting values, each number is the sum of the two preceding numbers. The first Fibonacci numbers (sequence A000045 in OEIS), also denoted as F
; 

F
can be calculate exactly by the following two expressions:





A Fibonacci spiral created by drawing arcs connecting the opposite corners of squares in the Fibonacci tiling; this one uses squares of sizes 1, 1, 2, 3, 5, 8, 13, 21, and 34;

So you can see how interesting the Fibonacci number is.

Now AekdyCoin denote a function G(n)



Now your task is quite easy, just help AekdyCoin to calculate the value of G (n) mod C

 

Input

The input consists of T test cases. The number of test cases (T is given in the first line of the input. Each test case begins with a line containing A, B, N, C (10<=A, B<2^64, 2<=N<2^64, 1<=C<=300)

 

Output

For each test case, print a line containing the test case number( beginning with 1) followed by a integer which is the value of G(N) mod C

 

Sample Input

1
17 18446744073709551615 1998 139

 

Sample Output

Case 1: 120根据题目给定的条件可以得到G（n）=F(a^b)^(F(a^b)^n-1)%c;根据公式：a^b%p=（a^(b%phi(p)+phi(p))）%p  进行降幂由于题目中C的范围比较小 因此我们一定可以找到它的循环节，进而求得F(a^b%c),F(a^b%phi(c));然后经过一系列的快速幂取模就可以得到答案

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;

typedef unsigned long long LL;

LL a,b,n;

int c;

int phi(int n){
int rea=n,i;
for(i=2;i*i<=n;i++){
if(n%i==0){
rea=rea-rea/i;
while(n%i==0)
n/=i;
}
}
if(n>1)
rea=rea-rea/n;
return rea;
}

LL multi(LL a,LL b,LL m)
{
LL ans=0;
while(b)
{
if(b&1)
{
ans=(ans+a)%m;
b--;
}
b>>=1;
a=(a+a)%m;
}
return ans;
}

LL quick_mod(LL a,LL b,LL m){
LL ans=1;
a%=m;
while(b){
if(b&1){
ans=multi(ans,a,m);
b--;
}
b>>=1;
a=multi(a,a,m);
}
return ans;
}

LL f[5000];

int find_loop(int c){
f[0]=0;f[1]=1;
int loop;
for(int i=2;i<5000;i++){
f[i]=(f[i-1]%c+f[i-2]%c)%c;
if(f[i]==1&&f[i-1]==0){
loop=i;
break;
}
}
return loop-1;
}
int main()
{
int t,tt=1;
LL tmp1,tmp2;
LL t1,t2;
scanf("%d",&t);
while(t--)
{
scanf("%I64u%I64u%I64u%d",&a,&b,&n,&c);
printf("Case %d: ",tt++);
if(c==1)
{
puts("0");
continue;
}
int p=phi(c);
int loop1=find_loop(c);
t1=quick_mod(a,b,loop1);
tmp1=f[t1]%c;
int loop2=find_loop(p);
t2=quick_mod(a,b,loop2);
tmp2=f[t2]%p;
tmp2=quick_mod(tmp2,n-1,p);
tmp2+=p;
tmp1=quick_mod(tmp1,tmp2,c);
printf("%I64u\n",tmp1);
}
return 0;
}
/*****
1
17 18446744073709551615 1998 139
Case 1: A  54
p    138
B   79
c    109
loop1    46
loop2    48
120

*****/