n=int(input())
l=[]
for i in range(n):
l.append(int(input()))
for i in range(0,2**n):
res=0
for j in range(n):
if(i &(2**j)):
res+=l[j]
else:
res-=l[j]
if res%360==0:
print("yes")
break
else:
print("no")
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<numeric>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<ll> vl;
#define all(v) v.begin(),v.end()
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
void Never_give_up() {
std::ios_base::sync_with_stdio(NULL);
cin.tie(NULL); cout.tie(NULL);
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif
}
void solve() {
int n; cin >> n;
bool Ok = false;
vi v(n); for (auto& it : v) cin >> it;
for (int mask = 0; mask < (1 << n); mask++) {
int sum = 0;
for (int i = 0; i < n; i++) {
if (mask & (1 << i)) sum += v[i];
else sum -= v[i];
}
Ok |= (sum % 360 == 0);
}
cout << ((Ok) ? "YES" : "NO");
}
int main() {
//وَلَسَوْفَ يُعْطِيكَ رَبُّكَ فَتَرْضَى
Never_give_up();
int T = 1; //cin >> T;
while (T--) {
solve();
}
return 0;
}
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490A - Team Olympiad | 233A - Perfect Permutation |
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