思路:問你放兩堆規模更小的石子,sg(bi,bj)=sg(bi)^sg(bj)
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define space putchar(' ')
#define enter putchar('\n')
typedef pair<int,int> PII;
const int inf=0x3f3f3f3f;
const int mod=1e9+7;
const int N=1e5+10;
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
ll lcm(ll a,ll b)
{
return a*(b/gcd(a,b));
}
template <class T>
void read(T &x)
{
char c;
bool op = 0;
while(c = getchar(), c < '0' || c > '9')
if(c == '-')
op = 1;
x = c - '0';
while(c = getchar(), c >= '0' && c <= '9')
x = x * 10 + c - '0';
if(op)
x = -x;
}
template <class T>
void write(T x)
{
if(x < 0)
x = -x, putchar('-');
if(x >= 10)
write(x / 10);
putchar('0' + x % 10);
}
int n,k;
int f[N],a[N];
int sg(int x)
{
if(f[x]!=-1)
return f[x];
unordered_set<int> S;
for(int i=0; i<x; i++)
{
for(int j=0;j<=i;j++)
{
S.insert(sg(i)^sg(j));
}
}
for(int i=0;; i++)
if(!S.count(i))
return f[x]=i;
}
int main()
{
memset(f,-1,sizeof f);
read(n);
int ans=0;
for(int i=0; i<n; i++)
{
int x;
read(x);
ans^=sg(x);
}
if(ans)
puts("Yes");
else
puts("No");
}