vjudge A ^ B Mod C and, most 10.6. . . Fast power! ! !

 

Links: https://vjudge.net/contest/331993#problem/D

 

3 gives a positive integer ABC, seeking A ^ B Mod C.

For example, 3 5 8,3 ^ 5 Mod 8 = 3.

Input3 positive integer ABC, separated by a space. (1 <= A, B, C <= 10 ^ 9) Output Sample Input outputs the calculation result

3 5 8

Sample Output

3

#include <cstdio> 
#include <the iostream>
 the using  namespace STD;
 int POW ( int A, int B, int C) 
{ 
    int ANS = . 1 , Base = A; // ANS: Results of power; base: base number A 
    the while ( b) 
    { 
        IF (& b . 1 )    // determination is not an odd b 
        { 
        ANS = ( Long  Long ) ANS * Base % C; 
        } 
        Base = ( Long  Long ) Base*base%c;
        b = b >> 1;  //除2 
    }
    return ans;
}
int main()
{
    int a,b,c;
    scanf("%d%d%d",&a,&b,&c);
    printf("%d\n",pow(a,b,c));
}

 

 

In fact, it can be used as a template to back off, -. - although I know it is. . .