6-3 Quick Power （10 分）

The function Power calculates the exponential function N​k​​. But since the exponential function grows rapidly, you are supposed to return (N​k​​)%10007instead.

Format of function:

int Power(int N, int k);

Both N and k are integers, which are no more than 2147483647.

Sample program of judge:

#include <stdio.h>int Power(int, int);const int MOD = 10007;
int main()
{int N, k;scanf("%d%d", &N, &k);printf("%d\n", Power(N, k));		return 0;
}/* Your function will be put here */

Sample Input 1:

2 3

Sample Output 1:

8

Sample Input 2:

128 2

Sample Output 2:

6377

 快速幂取模

//积的取余等于取余的积的取余
(a*b) % m = pi;
((a%m)*(b%m)) % m = pj;
即证：
pi = pj;
0 < pi < m, 0 < pj < m;
证明如下：
1.a = m*x1 + p1;
2.b = m*x2 + p2;
3.(a%m)*(b%m) = p1*p2;
4.(a%m)*(b%m) = m*x3 + pj;
5.a*b = m*x1*m*x2 + p1*m*x2 + p2*m*x1 + p1*p2
6.    = m*(m*x1*x2 + p1*m2 + p2*m1) + p1*p2;
7.a*b = m*x4 + pi;
8.6 & 3-- > a*b = m*(m*x1*x2 + p1*m2 + p2*m1) + (a%m)*(b%m);
9.4 & 8-- > a*b = m*(m*x1*x2 + p1*m2 + p2*m1 + x3) + pj;
10.9 & 7-- > m*(m*x1*x2 + p1*m2 + p2*m1 + x3) + pj = m*x4 + pi;
pj=pi,得证！
-->a的b次方 mod c = (a mod c)的b次方 mod c

//快速幂

直到b为1

具体操作代码如下

int Power(int N, int k) {int n = 1;N = N%MOD;while (k > 0) {if (k % 2 == 1)n = n*N%MOD;//当k为奇数时k /= 2;N = N*N%MOD;}return n;	
}