C++_（矩阵）快速幂

#include <iostream>
#include <cstdio>

using namespace std;
const int MOD = 1000;

int POW(long long &num);
int times;
long long N;

int main()
{
scanf("%d", ×);
while(times--)
{
scanf("%lld", &N);
printf("%d\n", POW(N));
}
return 0;
}

int POW(long long &num)
{
int ans = 1;
long long t = num;
while(t)
{
if(t&1)
{
ans = (ans*num)%MOD;
}
num = num*num%MOD;
t = t>>1;
}
return ans%10;
}

#include <iostream>
#include <cstdio>
#define MOD 10000

using namespace std;

struct Matrix
{
int lt, rt, lb, rb;
Matrix() : lt(1), rt(1), lb(1), rb(0){}
Matrix& operator*(const Matrix m)
{
Matrix temp = (*this);
lt = (temp.lt*m.lt)%MOD + (temp.rt*m.lb)%MOD;
rt = (temp.lt*m.rt)%MOD + (temp.rt*m.rb)%MOD;
lb = (temp.lb*m.lt)%MOD + (temp.rb*m.lb)%MOD;
rb = (temp.lb*m.rt)%MOD + (temp.rb*m.rb)%MOD;
return (*this);
}
};

long long n;

int getAns();

int main()
{
while(~scanf("%lld", &n))
{
if(n == -1) break;
if(n == 0)
{
printf("0\n");
continue;
}

printf("%d\n", getAns());

}
return 0;
}

int getAns()
{
Matrix m;
n--;
//long long t = n;
Matrix res;
Matrix temp;
while(n)
{
if(n&1)
{
m= m * res;
//cout << m.lt << " " << m.rt << endl << m.lb << " " << m.rb << endl;
}
res = res * res;
//cout << res.lt << " " << res.rt << endl << res.lb << " " << res.rb << endl;
n = n>>1;
}

return (m.rt)%MOD;
}