Fast Power modulo---> (a^b)%c

First you need to know a formula a*b%c= ((a%c) *b)%c, and then you can try to write an inefficient algorithm based on this formula.

/*     (a*a*a*...*a)%c  = ((a*a*). *a)%c*a)%c  = ((*.. *a)%c*a)%c*a)%c = (((((((a%c*a)%c...*a)%c*a)%c*a)%c*a  #include%c*/<stdio.h>int (int qmod a,int c) {    //According to the introduction of the formula, easy to know    int sum=a%c;    b=b-1;    while (b--)    {        sum=sum*a%c;    }    For more aesthetic simplicity, it can be written as    int sum=1;//a*1=a    while (b--)    {        sum=sum*a%c;    }    return sum;} int main () {//a*b%c== ((a%c) *b)%c  int a,b,c;  while (scanf ("%d%d%d", &a,&b,&c)!=eof)  {      printf ("%d\n", Qmod (A,b,c));}  }

The next step is to raise the efficiency of the problem, how to improve efficiency?

This formula A^b%c, which has been optimized for mod modulo

The key now is to optimize the a^b.

These two simple formulas need to be remembered: 1): a^b= (a^2) ^ (B/2)-->b for even 2): a^b= (a^2) ^ (B/2) *a-->b for Odd

#include <stdio.h>int qmod (int a,int b,int c) {    int sum=1;    while (b)    {        if (b&1)          sum=sum*a%c;//multiply by one A;        a=a*a%c;        b>>=1;    }    return sum;} int main () {//a*b%c== ((a%c) *b)%c  int a,b,c;  while (scanf ("%d%d%d", &a,&b,&c)!=eof)  {      printf ("%d\n", Qmod (A,b,c));}  }

Attached: 1): YY right Shift x represents yy divided by 2^x 2):yy left Shift x represents yy multiplied by 2^x

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Fast Power modulo---> (a^b)%c