leetcode+pow函数实现，快速幂。可以重载leetcode函数，trick

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<cstring>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include <limits.h>
#include <assert.h>
#define LL long long
#define inf 0x3f3f3f3f
#define mod 1e9+7
using namespace std;
class Solution {
public:
double myPow(double x, int n) {
double ans=1;
unsigned long long p;
if(n<0){
p=-n;
x= 1/x;
}
else p=n;
while (p) {
if(p&1) ans*=x;
x*=x;
p=p>>1;
}
return ans;
}
};
int main()
{
Solution temple;
cout<<temple.myPow(2, -1)<<endl;
return 0;
}

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string>
#include<cstring>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include <limits.h>
#include <assert.h>
#define LL long long
#define inf 0x3f3f3f3f
#define mod 1e9+7
using namespace std;
//位运算基本上是不需要用的
class Solution {
public:
double myPow(double x, long long n) {//重载函数
double ans=1;
if(n<0){
x= 1/x;
n=-n;
}
while (n) {
if(n&1) ans*=x;
x*=x;
n=n/2;
}
return ans;
}
};
int main()
{
Solution temple;
cout<<temple.myPow(1, -2147483648)<<endl;
return 0;
}