23 lines
616 B
C++
23 lines
616 B
C++
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#include <bits/stdc++.h>
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using namespace std;
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// https://www.jianshu.com/p/749b6ee1d001
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//栗子： 2^11 = 2 ^ (1011)2
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//就是把11拆成二进制数位，然后从右到左进行次方base的迭乘
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//在从右到左的枚举过程中，发现本位是1,就乘上本位的base
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//原来需要2*2*...*2，共10次，即O(N)
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//现在需要log以2为底11就是log_2{11} 即4次
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int qmi(int a, int k) {
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int ans = 1;
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while (k) {
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if (k & 1) ans *= a;
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a *= a; //次方base
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k >>= 1;
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}
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return ans;
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}
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int main() {
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cout << qmi(2, 11) << endl;
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return 0;
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}
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