The power of dichotomy

int getMi(int a,int b)
int ans = ;
while (b != )
// When the binary bit k Position as 1 when , It's a long ride a Of 2^k Power , And then use ans preservation
if (b % == )
ans *= a;
a *= a;
b /= ;
return ans;

Fast modular exponentiation

The formula :

The final version of the algorithm :

int PowerMod(int a, int b, int c)
int ans = ;
a = a % c;
if(b % = = )ans = (ans * a) % c;
b = b/;
a = (a * a) % c;
return ans;

seek Root(N,k)

Title Description
       N<k when ,root(N,k) = N, otherwise ,root(N,k) = root(N',k).N' by N Of k The sum of the digits in decimal notation . Input x,y,k, Output root(x^y,k) Value ( here ^ For the power of , It's not XOR ),2=<k<=16,0<x,y<2000000000, Half of the test sites x^y It will overflow int The scope of the (>=2000000000) 
Input description
       Each set of test data consists of one line ,x(0<x<2000000000), y(0<y<2000000000), k(2<=k<=16)
Output description      
  Input may have multiple sets of data , For each set of data ,root(x^y, k) Value 
4 4 10

Code :

#include <cstdio>
#include <math.h>
#include <cstring>
#include <algorithm> //root(x*y,k) = root(root(x,k)*root(y,k),k) int Root(int N,int k)
if(N<k)return N;
int ans = ;
//N Greater than k, seek N by k The sum of all of you
while(N != ){
ans += N%k;
N /= k;
return Root(ans,k);
} int getAns(int x,int y,int k)
int num = Root(x,k);
int ans = ;
while(y > ){
if(y%){//y It's odd
ans = Root(ans*num, k);
y /= ;
num = Root(num*num, k);
return ans;
} int main()
int x,y,k; while(~scanf("%d %d %d",&x,&y,&k)){
printf("%d\n",getAns(x,y,k)); } return ;

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