Codeforces Round #670 (Div. 2) B. Maximum Product (暴力)

• 题意:有一长度为\(n\)的序列,求其中任意五个元素乘积的最大值.

• 题解:先排序,然后乘积能是正数就搞正数,模拟一下就好了.

• 代码:

  int t;
  ll n;
  ll a[N];
  
  int main() {
      ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
      cin>>t;
      while(t--){
      	cin>>n;
      	for(int i=1;i<=n;++i){
      		cin>>a[i];
      	}
      	sort(a+1,a+1+n);
      	ll cnt1=a[n]*a[n-1]*a[n-2]*a[n-3]*a[n-4];
      	ll cnt2=a[1]*a[2]*a[3]*a[4]*a[5];
      	ll cnt3=a[n]*a[1]*a[2]*a[3]*a[4];
      	ll cnt4=a[1]*a[2]*a[n]*a[n-1]*a[n-2];
      	ll res=max(cnt1,max(cnt2,max(cnt3,cnt4)));
      	cout<<res<<endl;
      }
  
      return 0;
  }