57 lines
1.0 KiB
C++
57 lines
1.0 KiB
C++
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#include<iostream>
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using namespace std;
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/*
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The idea is to half the power on every iteration
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Computes the power in log(n) operations
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On every iteration:
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Square the base, half the power
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Special case - if power is odd:
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Multiply the ans with a
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Because the ODD-NUMBER % 2 == 1
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Note: There will always be one iteration where power is odd
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*/
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long binpow(long a, long b){
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long ans=1;
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// while b is greater than zero, we continue the binary exponentiation
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while(b>0){
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// if b is odd, multiply result with base
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if(b&1)
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ans *= a;
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// square the base
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a *= a;
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// half the power
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b /= 2;
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}
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return ans;
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}
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// source: @angshudas
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long binpow_rec(long a, long b){
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// a^0 = 1 [for any a]
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if(b==0)
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return 1;
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// recursive call for a^(b/2)
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long ans = binpow_rec(a, b/2);
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// square the result
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ans *= ans;
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// if b is odd, times by a
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// to cover for
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if(b&1)
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ans *= a;
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return ans;
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}
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int main(){
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long base, power;
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cout<<"Enter the base and power: "<<endl;
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cin>>base>>power;
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cout<<base<<" ^ "<<power<<" = "<<binpow(base, power)<<endl;
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return 0;
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}
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