1600: Big Mod (Modulo of large numbers)

Result	Time Limit	Memory limit	Run times	AC times	Judge
	3 S	8192 K	1756	336	Standard

Calculate

 

 

For large valuesB,P, AndMUsing an efficient algorithm. (That's right, this problem has a time dependency !!!.)

Input

Three integer values (in the orderB,P,M) Will be read one number per line.BAndPAre integers in the range 0 to 2147483647 random Sive.MIs an integer in the range 1 to 46340 random Sive.

Output

The result of the computation. A single integer.

Sample Input

 

3181321717176532374859302938236123

Sample output

 

13213195/*

*/

1. # Include <stdio. h>
2. # Include <memory>
3. Int main ()
4. {
5. Int A, B, N;
6. While (scanf ("% d", & A, & B, & N )! = EOF)
7. {
8. Bool B2 [35];
9. Memset (B2, 0, sizeof (B2 ));
10. Int TMP = B, I, D = 1;
11. For (I = 0; TMP> 0; I ++)
12. {
13. B2 [I] = TMP % 2;
14. TMP = tmp/2;
15. }
16. For (; I> = 0; I --)
17. {
18. D = D * D % N;
19. Int AA = A % N;
20. If (B2 [I] = 1)
21. D = D * Aa % N;
22. }
23. Printf ("% d/N", d );
24. }
25. Return 0;
26. }

Method 2: ultra-short code

# Include <cstdio>
Int B, P, M, K;
Int main ()
{
While (scanf ("% d", & B, & P, & M )! = EOF)
{
For (k = 1, B = B % m; P/= 2, B = B * B % m)
If (P % 2) K = K * B % m;
Printf ("% d \ n", K % m );
}
Return 0;
}

 

Method 3: lower recursion time

# Include <cstdio>
Int m;
Int bigmod (int A, int B)
{
If (! B) return 1;
Int d = bigmod (a, B> 1) % m;
Return B & 1? (D * D % M * A) % m :( D * D) % m;
}
Int main ()
{
Int A, B;
While (scanf ("% d", & A, & B, & M )! = EOF)
{
Int ans = bigmod (a % m, B) % m;
Printf ("% d \ n", ANS );
}
Return 0;
}

The shortest code for rapid modulo operation (k = 1, A = A % m; B; B >>= 1, A = A * A % m) if (B & 1) k = K * A % m; find a ^ B (mod m) K % m as the answer.