P5514 [MtOI2019]永夜的报应

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;
}