uva 11582(大fib，打表找循环节）

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

Sample input

three integers a,b,n where 0 ≤ a,b < 264 (a and b will
not both be zero) and 1 ≤ n ≤ 1000.

T

a  b  n 

3
1 1 2
2 3 1000
18446744073709551615 18446744073709551615 1000

Sample output

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

1
21
250

其实这题的主要知识，还是找循环节，打表。

卡点：2^64-1 --->   unsigned long long

#include<iostream>
#include<cstdio>
#include<cmath>
#include<vector>
#define bug(a) cout<<a<<"--->\n";
using namespace std;

typedef unsigned long long ULL;

vector<int>f[1005];
int period[1005];

int qpow(ULL a,ULL b,int p)
{
int ans=1;
while(b)
{
if(b&1) ans=int((ans*a)%p);
a=(a*a)%p;
b>>=1;
}
return ans;
}

void pre_fib()
{
for(int n=2;n<=1000;n++)
{
f
.push_back(0);f
.push_back(1);
for(int i=2;;i++)
{
f
.push_back((f
[i-1]+f
[i-2])%n);
if(f
[i-1]==0&&f
[i-2]==1)
{
period
=i-1;
break;
}
}
}
}

int main()
{
int T;
pre_fib();
scanf("%d",&T);
while(T--)
{
ULL a,b;
int p;
int c;
scanf("%llu%llu%d",&a,&b,&p);
if(a==0||p==1) printf("0\n");
else
{
c=qpow(a%period[p],b,period[p]);
printf("%d\n",f[p][c]);
}
}
return 0;
}