排个序就过了，需要XOR？

#include <iostream>
#include <algorithm>

using namespace std;

int n;
long long num[1000000]={0};

int main()
{
    cin>>n;
    int i;
    for(i=0;i<n;i++)
    {
        cin>>num[i];
    }
    sort(num,num+n);
    for(i=0;i<n;i+=2)
    {
        if(num[i]!=num[i+1])
        {
            cout<<num[i]<<endl;
            break;
        }
    }
    return 0;
}

你确定这是hash？？？

var
  s, t:qword;
  n, i:longint;
begin
  readln(n);
  s:=0;
  for i:=1 to n do begin
    read(t);
    s:=s xor t
  end;
  write(s)
end.

居然要输入挂了

c++两行水过
#include<iostream>
using namespace std;long long a,b,c;main(){cin>>a;while(a--)cin>>b,c^=b;cout<<c;}

program pp;
var
i,k:longint;
a,j:int64;
begin
readln(k);
for i:=1 to k do
begin
read(j);
a:=a xor j;
end;
write(a);
end.

program p1684;
var i,j,k,m,n:longint;
d,max:int64;
begin
readln(k);
for i:=1 to k do
begin
read(d);
max:=max xor d;
end;
writeln(max);
end.

不给范围差评，裸map过

#include <iostream>
#include <map>
#include <cstdio>
using namespace std;

map<long long, long long> biao;
int n;

int main()
{
scanf ("%d", &n);
for (int i=1; i<=n; i++) {
long long temp;
scanf ("%lld", &temp);
if (!biao.count(temp)) biao[temp] = 0;
biao[temp]++;
}
for (map<long long, long long>::iterator i=biao.begin(); i!=biao.end(); ++i) {
if (i->second & 1) {
cout << i->first << endl;
return 0;
}
}
return 0;
}

#include<cstdio>
using namespace std;

long long int ans=0, l;

int main()
{
int n;
scanf("%d",&n);
for(int i=1; i<=n; i++){
scanf("%lld",&l);
ans ^= l;
}
printf("%lld",ans);

return 0;
}
水一发~~~

为何这样不行？卡在四号上了。
Program P1684;
Const MAX=10000000;
Var bl:array[1..MAX]of boolean;
n,o,i:longint;
begin
readln(n);
For i:=1 to n do
begin
read(o);
bl[o]:=not bl[o];
end;
For i:=1 to MAX do
If bl[i] then break;
writeln(i);
end.
最后还是用了xor。。。

此题数据范围略大，要用64位整形储存，全部xor一边后即是答案

c++ code

#include<cstdio>
int main()
{
long long n,ans=0,t;
scanf("%lld",&n);
for (long long i=0;i!=n;++i)
{ scanf("%lld",&t); ans^=t; }
printf("%lld",ans);
return 0;
}

program P1684;
var ll,l:int64;
i,n:longint;
begin
read(n);
for i:=1 to n do
begin
read(l);ll:=ll xor l;
end;
write(ll);
end.

#include<cstdio>
#define IOFileName "P1684"
int n;
unsigned long long A=0,B;
int main(){
//freopen(IOFileName".in","r",stdin);
//freopen(IOFileName".out","w",stdout);
scanf("%d\n",&n);
for(int i=0;i<n;i++){
scanf("%I64u ",&B);
A^=B;
}
printf("%I64u\n",A);
}
357MS+248KB

测试数据 #0: Accepted, time = 0 ms, mem = 604 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 608 KiB, score = 10

测试数据 #2: Accepted, time = 31 ms, mem = 608 KiB, score = 20

测试数据 #3: Accepted, time = 0 ms, mem = 612 KiB, score = 20

测试数据 #4: Accepted, time = 202 ms, mem = 608 KiB, score = 40

Accepted, time = 233 ms, mem = 612 KiB, score = 100

耗时233纪念，学习位运算的重要性2333333333

我去！一水题废我过题率！var
j,n,m:int64;
i:longint；
begin
{assign(input,'sort.in');
assign(output,'sort.out');
reset(input);
rewrite(output);}
readln(n);
read(m);
for i:=1 to n-1 do
begin
read(j);
m:=m xor j;
end;
writeln(m);
{ close(input);
close(output); }
end.

var
n,i:longint;
x,y:int64;
begin
readln(n);
read(x);
for i:=2 to n do
begin
read(y);
x:=x xor y;
end;
writeln(x);
end.
详见百度位运算

这题百度百科位运算里面有的，神奇的XOR= =你值得拥有

var a:array[1..10000000] of int64;
i,j,k,l,q,w,e,n,m:longint;
procedure qsort(l,r:longint);
var i,j,mid,e:int64;
begin
i:=l;j:=r;mid:=a[random(r-l+1)+l];
repeat
while a[i]<mid do inc(i);
while a[j]>mid do dec(j);
if i<=j then begin e:=a[i];a[i]:=a[j];a[j]:=e;inc(i);dec(j);end;
until i>j;
if i<r then qsort(i,r);
if j>l then qsort(l,j);
end;
begin
readln(n);
for i:=1 to n do read(a[i]);
qsort(1,n);
for i:=1 to n div 2+1 do
if a[i*2]<>a[i*2-1] then begin writeln(a[i*2-1]);halt;end;
end.

快排优化AC过

本题提交2100次纪念。

(a xor b)xor b=a即为本题的精髓所在。

program p1684(input,output);

var n,zs,zhi:int64;

i:longint;

begin

readln(n);

zs:=0;

for i:=1 to n do

begin

read(zhi);

zs:=zs xor zhi;

end;

writeln(zs);

end.

闪过……………………

/*

Vijos P1684

Time 2010/02/28

Lang C

*/

#include

main()

{

long long i,j,k,ans;

scanf("%I64d",&i);

ans = 0;

for (j=1;j