题解 | #日渐圆润的川页姐#

第三道题是二进制运算 与有关快速幂

#include<bits/stdc++.h>
#define ll long long
#define endl '\n'
using namespace std;
void solve()
{	
//记得开long long 此处宏定义ll为long long
	ll k;cin>>k;
	ll ans=0;
	ll weight=1;
	while(k){
		if(k&1){//与1,能够判断最后一位是不是1
			ans+=weight;	
		}
		k>>=1;//然后位运算右移
		weight*=3;
	}
	cout<<ans<<endl;
	
}

int main()
{
    //关闭输入输出流同步 
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    //测试样例 
    ll t;
    cin>>t;
    while(t--)
    {
	   solve();
    }
}

附快速幂的模板

#include<bits/stdc++.h>
#define ll long long
#define endl '\n'
using namespace std;
//快速幂模板,二进制运算 
long long fastPower(long long base, long long power) {
    long long result = 1;
    while (power > 0) {
        if (power & 1) {//此处等价于if(power%2==1)
            result = result * base % 1000;
        }
        power >>= 1;//此处等价于power=power/2
        base = (base * base) % 1000;
    }
    return result;
}
void solve(){
	int zs,ds;
	cin>>ds;
	cin>>zs;
	cout<<fastPower(ds,zs)<<endl;
	
}
int main(){
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	cin>>t;
	while(t--){
		solve();
	}
	
}