Codeforces 429D Tricky Function

Codeforces 429D Tricky Function

This is equivalent to I representing the X axis, sum [I] representing the Y axis, sum [I] representing the sum of the previous I numbers

Calculate the distance between the minimum two points.

The so-called question is the question that suddenly becomes enlightened after reading the question, that is, the distance between j-> I first calculated during calculation,

# Include

 
  
# Include

  
   
# Include

   
    
Using namespace std; typedef long ll; # define N 11111ll a [N]; ll sum [N]; int ans; int main () {int n; sum [0] = 0; scanf ("% d", & n); for (int I = 1; I <= n; I ++) {scanf ("I64d", & a [I]); sum [I] = sum [I-1] + a [I];} ans = a [2] * a [2] + 1; for (ll I = 1; I

    
     
Ans) break; // so importantll t = ans; for (int j = I + 1; j <= n; j ++) {t = min (t, (sum [j]-sum [j-I]) * (sum [j]-sum [j-I]);} ans = min (ans, t + I * I);} cout <