非本人所写，作
program e1201;
var n,m,k,l,x,y:int64;
i,j:longint;
a,b,c,f:array[0..100]of int64;
begin
readln(n);
k:=0;
for i:=1 to 16 do
begin
a[i]:=n and 1;
n:=n shr 1;
end;
for i:=1 to 16 do
begin
b[i]:=n and 1;
n:=n shr 1;
end;
for i:=16 downto 1 do
begin
inc(k);
c[k]:=a[i];
end;
for i:=16 downto 1 do
begin
inc(k);
c[k]:=b[i];
end;
for i:=1 to k do
if c[i]=1 then begin
inc(l);
f[l]:=32-i;
end;
for i:=1 to l do
begin
y:=1;
for j:=1 to f[i] do
y:=y*2;
x:=x+y;
end;
writeln(x);
end.

program e1201;
var n,m,k,l,x,y:int64;
i,j:longint;
a,b,c,f:array[0..100]of int64;
begin
readln(n);
k:=0;
for i:=1 to 16 do
begin
a[i]:=n and 1;
n:=n shr 1;
end;
for i:=1 to 16 do
begin
b[i]:=n and 1;
n:=n shr 1;
end;
for i:=16 downto 1 do
begin
inc(k);
c[k]:=a[i];
end;
for i:=16 downto 1 do
begin
inc(k);
c[k]:=b[i];
end;
for i:=1 to k do
if c[i]=1 then begin
inc(l);
f[l]:=32-i;
end;
for i:=1 to l do
begin
y:=1;
for j:=1 to f[i] do
y:=y*2;
x:=x+y;
end;
writeln(x);
end.

###上面的抄我程序干嘛
测试数据 #0: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #1: Accepted, time = 11 ms, mem = 532 KiB, score = 10
测试数据 #2: Accepted, time = 15 ms, mem = 532 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 524 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 528 KiB, score = 10
测试数据 #6: Accepted, time = 1 ms, mem = 528 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 528 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 528 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 528 KiB, score = 10
Accepted, time = 57 ms, mem = 532 KiB, score = 100
#include <iostream>
#include <stdio.h>
using namespace std;
int num2[33];
int num_2[33];
int main()
{
int n;
scanf("%d",&n);
for(int i=1;;i++)
{
if(n==0)break;
if(n%2==1)
num2[i]=1;
n/=2;
}
for(int i=1;i<=16;i++)
{
num_2[i]=num2[i+16];
num_2[i+16]=num2[i];
}
long long ans=0;
long long n_2=1;
for(int i=1;i<=32;i++)
{
if(num_2[i])ans+=n_2;
n_2*=2;
}
printf("%lld",ans);
}

var
s,q:qword;
a:array[1..32] of longint;
i:longint;
begin
read(s);
i:=32;
while s<>0 do
begin
a[i]:=s mod 2;
s:=s div 2;
i:=i-1;
end;
q:=1;
for i:=16 downto 1 do
begin
s:=s+a[i]*q;
q:=q*2;
end;
for i:=32 downto 17 do
begin
s:=s+a[i]*q;
q:=q*2;
end;
write(s);
end.

swap函数可以直接用。

Pascal Code

var
s:longword;
begin
readln(s);
writeln(swap(s));
end.

#include <iostream>
using namespace std;
int main()
{
unsigned int n=0;
cin>>n;
cout<<(n>>16)+(n<<16)<<endl;
return 0;
}

P1201高低位交换
Accepted

记录信息

评测状态 Accepted
题目 P1201 高低位交换
递交时间 2014-11-05 02:06:49
代码语言 C++
评测机 上海红茶馆
消耗时间 7 ms
消耗内存 540 KiB
评测时间 2014-11-05 02:06:50

评测结果

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 540 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 532 KiB, score = 10

Accepted, time = 0 ms, mem = 540 KiB, score = 100

代码

#include <stdio.h>

unsigned int n;

int main()
{
scanf( "%u" , &n );
printf( "%u\n" , n >> 16 | n << 16 );
return 0;
}

左右位移。。。

#include <cstdio>
int main(void){
unsigned long x;
scanf("%lu\n",&x);
unsigned long y=x>>16|x<<16;
printf("%lu\n",y);
}
70分的是因为最前面一位被当成了符号位，输出时有负数。把％ld换成％lu就可以了

此题神水啊！
var
x:qword;
begin
readln(x);
writeln(x mod (1<<16)*(1<<16)+x>>16);
end.

var i,j,k,l,m,n:Longint;
NN,Ans:Int64;
S:string;
Begin
Readln(NN);
For i:=1 to 32 do
Begin
S:=chr(NN mod 2+48)+S;
NN:=NN div 2;
End;
S:=Copy(S,17,16)+Copy(S,1,16);
For i:=1 to 32 do Ans:=Ans*2+ord(S[i])-48;
Writeln(Ans);
End.

我累个擦，longint改成longword就过，没天理

我爱老夏!!老夏爱XX!!!>.

记得 int64

坑爹啊~pascal竟然还有swap函数

点击查看代码[/red]

编译通过...

├ 测试数据 01：答案正确... 0ms

├ 测试数据 02：答案正确... 0ms

├ 测试数据 03：答案正确... 0ms

├ 测试数据 04：答案正确... 0ms

├ 测试数据 05：答案正确... 0ms

├ 测试数据 06：答案正确... 0ms

├ 测试数据 07：答案正确... 0ms

├ 测试数据 08：答案正确... 0ms

├ 测试数据 09：答案正确... 0ms

├ 测试数据 10：答案正确... 0ms

---|---|---|---|---|---|---|---|-

Accepted 有效得分：100 有效耗时：0ms

强势位运算~！

Flag 　　Accepted

题号 　　P1201

类型(?) 　　数论 / 数值

通过 　　3946人

提交 　　7269次

通过率 　　54%

难度 　　1

#include

#include

int main()

{

long long n,m;

long long x=0,y=0;

scanf("%I64d",&n);

x=n>>16;

y=n-((n>>16)

个人觉得还是这个比较简洁

var a:dword;

begin

read(a);

a:=a shr 16+a shl 16;

write(a);

end.

貌似还有很多办法，位运算应该最优了。

第30个AC...小小小小庆祝

现在终于知道还有类型是

dword longword

还有函数是

swap