题解 | #2的n次方计算#
2的n次方计算
https://www.nowcoder.com/practice/35a1e8b18658411388bc1672439de1d9
#include <iostream> using namespace std; int main() { int n; int sum=1; cin>>n; //for(int i=0;i<n;i++){ // sum*=2; //} cout<<(1<<n); }
1<<n 就是2的n次方
2的n次方计算
https://www.nowcoder.com/practice/35a1e8b18658411388bc1672439de1d9
#include <iostream> using namespace std; int main() { int n; int sum=1; cin>>n; //for(int i=0;i<n;i++){ // sum*=2; //} cout<<(1<<n); }
1<<n 就是2的n次方
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