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
Input begins with an integer t ≤ 10, 000, the number of test cases.
Each test case consists of three integers a, b, n where 0 ≤ a, b < 2 64 (a and b will not both be zero) and 1 ≤ n ≤ 1000.
Output
For each test case, output a single line containing the remainder of f(a b ) upon division by n.
Sample Input
3 1 1
2 2 3
1000 18446744073709551615 18446744073709551615 1000
Sample Output
1
21
250
题解：
模板题。
AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const int maxn=10000000+5;
typedef unsigned long long ll;
int s[maxn];
ll pow_mod(ll a,ll b,int n)
{
a%=n;
ll ans=1;
while(b)
{
if(b&1)
ans=ans*a%n;
b=b>>1;
a=a*a%n;
}
return ans;
}
int main()
{
int num;
while(cin>>num)
{
while(num--)
{
ll a,b;
int n;
cin>>a>>b>>n;
if(n==1||a==0)
{
cout<<"0"<<endl;
continue;
}
s[0]=s[1]=1;
int p;
for(int i=2;;i++)
{
s[i]=(s[i-1]+s[i-2])%n;
if(s[i]==1&&s[i-1]==1)
{
p=i-1;
break;
}
}
cout<<s[pow_mod(a,b,p)-1]<<endl;
}
}
return 0;
}
今天也是元气满满的一天！good luck！