水题: 费马小定理+快速幂+矩阵快速幂
(第一次用到费马小定理)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL MOD = 1000000006;
const LL MOD1 = 1000000007;
struct Matrix
{
LL NUM[2][2];
Matrix operator + (const Matrix a) const
{
Matrix c;
for(int i = 0; i < 2; ++i)
{
for(int j = 0; j < 2; ++j)
{
c.NUM[i][j] = NUM[i][j] + a.NUM[i][j];
}
}
return c;
}
Matrix operator * (const Matrix a) const
{
Matrix c;
for(int i = 0; i < 2; ++i)
{
for(int j = 0; j < 2; ++j)
{
c.NUM[i][j] = 0;
for(int k = 0; k < 2; ++k)
c.NUM[i][j] = (c.NUM[i][j] + NUM[i][k] * a.NUM[k][j] % MOD) % MOD;
}
}
return c;
}
}; Matrix ppow(Matrix a, LL n)
{
Matrix ret;
for(int i =0 ; i< 2; ++i)
{
for(int j = 0; j < 2; ++j)
ret.NUM[i][j] = i==j ? 1 : 0;
}
while(n)
{
if(n & 1) ret = ret * a;
a = a * a;
n >>= 1;
}
return ret;
} LL Pow(LL a, LL n)
{
LL ret = 1;
while(n)
{
if(n & 1) ret =ret * a % MOD1;
a = a * a % MOD1;
n >>= 1;
}
return ret;
} int main()
{
LL a, b, n;
Matrix E;
E.NUM[0][0] = 1; E.NUM[0][1] = 1;
E.NUM[1][0] = 1; E.NUM[1][1] = 0;
while(cin >> a >> b >> n)
{
if(n == 0) cout << a << endl;
else if(n == 1) cout << b << endl;
else
{
n -= 1;
Matrix tmp = ppow(E,n);
LL na = tmp.NUM[0][1] , nb = tmp.NUM[0][0];
LL ans = (Pow(a,na) * Pow(b,nb))%MOD1;
cout << ans << endl;
}
}
return 0;
}