异或是不进位的加法，即 (a xor b le a+b)
所以把所有数字异或起来就好
my code:
#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i(j);i<=k;++i)
#define drp(i,j,k) for(register int i(j);i>=k;--i)
using namespace std;
typedef long long lxl;
template<typename T>
inline T max(T &a, T &b) {
return a > b ? a : b;
}
template<typename T>
inline T min(T &a, T &b) {
return a < b ? a : b;
}
inline char gt() {
static char buf[1 << 21], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T>
inline void read(T &x) {
register char ch = gt();
x = 0;
int w(0);
while(!(ch >= '0' && ch <= '9'))w |= ch == '-', ch = gt();
while(ch >= '0' && ch <= '9')x = x * 10 + (ch & 15), ch = gt();
w ? x = ~(x - 1) : x;
}
template <typename T>
inline void out(T x) {
if(x < 0) x = -x, putchar('-');
char ch[20];
int num(0);
while(x || !num) ch[++num] = x % 10 + '0', x /= 10;
while(num) putchar(ch[num--]);
putchar('
');
}
int n;
int ans,a;
int main() {
read(n);
read(ans);
rep(i,2,n){
read(a);
ans^=a;
}
out(ans);
return 0;
}