#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
const int INF=0x3f3f3f3f;
unsigned int n,a;
int main(){
	cin>>n;
	a=(n>>16)+(n<<16);
	cout<<a;
	
	return 0;
}
小鸟

#include<bits/stdc++.h>
using namespace std;
unsigned int n,a;
int main(){
	cin>>n;
	a=(n>>16)+(n<<16);
	cout<<a;
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=1e3+10;
const int INF=0x3f3f3f3f;
unsigned int n,a;
int main(){
	cin>>n;
	a=(n>>16)+(n<<16);
	cout<<a;
	
	return 0;
}

#include<cstdio>
#include<string>
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
const int N=2e5+10;
int n,x;
int main(){
	unsigned int a;
	cin>>a;
	cout<<(a>>16)+(a<<16);
	return 0;
}

···#include<iostream> #include<stdio.h> #include<math.h> #include<iomanip> #include<string.h> #include<algorithm> #define LL long long using namespace std; const int N=1e5+10; const int INF=0x3f3f3f3f; int main(){ long long a,b; cin>>a; LL c=a>>16; b=a-(c<<16); LL x=(b<<16)+c; cout<<x; }

1.int型最多存储31位整形[即正负数都有，范围在-$2^{31}$ − $2^{31}$-1]，而unsigned int也是31位，但是存储的是正整数[范围是$1$ − $2^{32}$-1]。

2.C++运算中，若出现溢出，则 自动取模 。举个例子：在unsigned int型当中，计算$2^{32}$+400时，实际的运算是($2^{32}$+400)%$2^{32}$，结果就是400.

综合以上几点，能不能发现什么呢？

问题让我们求的其实是2进制下交换前后16位，不就是向左移16位（根据上文2.），加上右移16位吗？

代码如下

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
    unsigned int n;
    cin>>n;
    cout<< (n<<16)+(n>>16);
    return 0;
}

#include<iostream>
#include<cmath>
using namespace std;
string t00(int num){
	string s = "";
	while(num){
		s += char(num%2+'0');
		num /= 2;
	}
	for(int i = 0; i < s.size() / 2; i++){
		swap(s[i],s[s.size()-i-1]);
	}
	if(s.size() < 32){
		while(s.size() != 32){
			s = '0' + s;
		}
	}
	return s;
}
string sw(string s){
	for(int i = 0; i < 16; i++){
		swap(s[i],s[i+16]);
	}
	return s;
}
long long t01(string s){
	long long ans = 0;
	for(int i = 31; i >= 0; i--){
		ans += (long long)((s[i]-'0') * pow(2,31-i));
	}
	return ans;
}
void run(){
	int n;
	cin >> n;
	cout << t01(sw(t00(n)));
}
int main(){
	run();
}

#include<bits/stdc++.h>
const int N=1e5+10;
const int INF=0x3f3f3f3f;
unsigned int n,a;
using namespace std;
int main(){
	cin>>n;
	a=(n>>16)+(n<<16);
	cout<<a;
	return 0;
}

#include<iostream>
#include<cstring>
using namespace std;
unsigned int n,sum; 
int main(){
	cin>>n;
	sum=(n>>16)+(n<<16);
	cout<<sum;
} 

#include<bits/stdc++.h> using namespace std; int main() { //int x,s,n //低位转换 //高位转换 //输出 unsigned x,s,n; cin>>n; s=(n<<16); x=(n>>16); cout<<s+x<<endl; }

#include<bits/stdc++.h> using namespace std; int main() { //int x,s,n //低位转换 //高位转换 //输出 unsigned x,s,n; cin>>n; s=(n<<16); x=(n>>16); cout<<s+x<<endl; }

#include<bits/stdc++.h> using namespace std; int main() { //int x,s,n //低位转换 //高位转换 //输出 unsigned x,s,n; cin>>n; s=(n<<16); x=(n>>16); cout<<s+x<<endl; }

#include<bits/stdc++.h> using namespace std; int main() { unsigned n; int a,x; cin>>n; cout<<(n>>16)+(n<<16)<<endl; return 0; }

using namespace std;
unsigned n;
int main(){
	cin >> n;
	cout << (n >> 16)+(n << 16);
	return 0;
}//超简单的，不需要太多

#include <queue>
#include <math.h>
#include <stack>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <iomanip>
#include <string.h>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
int main()
{
//  unsigned int a,b;
//   cin >> a;
//  unsigned int c=a>>16;
//  b=a-(c<<16);
//  unsigned int x=(b<<16)+c;
//  cout << x <<endl;
    unsigned int a;
    cin >> a;
	cout << (a << 16)+(a >> 16);
}

#include <math.h>
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
int main()
{
    long long  n ;
    cin >> n;
    long long h = n >> 16;
    n = n - (h << 16);
    long long sum = (n << 16) + h;
    cout << sum << endl;
    return 0;
}