27 lines
469 B
C++
27 lines
469 B
C++
#include <bits/stdc++.h>
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using namespace std;
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int b, p, k;
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/*
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原理：a*b%k=(a%k)*(b%k)%k
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对于任何一个自然数：p=2*(p/2)+p%2
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*/
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int f(int p) {
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if(p==0) // b^0%k
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return 1%k;
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int t=f(p/2)%k;
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t=(t*t)%k; // b^p%k=(b^(p/2))^2%k
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if(p%2==1)
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t=(t*b)%k; // 如果p为奇数，则b^p%k=((b^(p/2))^2)*b%k
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return t;
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}
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int main(){
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cin>>b>>p>>k;
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int tmpb=b;
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b%=k;
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printf("%d^%d mod %d=%d", tmpb, p, k, f(p));
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return 0;
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}
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