#include<bits/stdc++.h> //#include "squares.h" using namespace std; vector<int> paint(int n){ //low n test cases if (n<=1000){ vector<int> ans(n+1,0); ans[n]=n; return ans; } //the actual solution vector<int> v,vv; for (int bit=0; bit<67 && v.size()!=n; bit++){ vv.clear(); for (int i=0; i<7; i++){ int tr = (bit >> i) & 1; vv.push_back(tr); } for (int j=6; j>=0 && v.size()!=n; j--){v.push_back(vv[j]);} for (int j=8; j>=0 && v.size()!=n; j--){v.push_back(1);} } v.push_back(16); return v; } int find_location(int n, vector<int> vx){ //low n test cases if (n<=1000){ int ans=0; for (auto x : vx){if (x==-1){ans++;}} return ans; } //ending with -1 int x=n-16; for (int i=n-1; i>=0; i--){ if (vx[i]==-1){ x++; } } if (x>n-16){return x;} //the actual solution vector<int> v,vv; for (int bit=0; bit<67 && v.size()!=n; bit++){ vv.clear(); for (int i=0; i<7; i++){ int tr = (bit >> i) & 1; vv.push_back(tr); } for (int j=6; j>=0 && v.size()!=n; j--){v.push_back(vv[j]);} for (int j=8; j>=0 && v.size()!=n; j--){v.push_back(1);} } for (int i=0; i<n-15; i++){ vv.clear(); for (int j=i; j<i+16; j++){ vv.push_back(v[j]); } if (vv==vx){return i;} } return 0; } // subtask 1,2,3 -- > 70 = 2 * 35 -- > ((35 * 36 ) / 2) *2 > 1000 hence we can just do 35 1 then 35 0 then 34 1 ... so on // subtask 4,5 {half point} 2^10 > 1024 {xxxxxxxxxx011111111}; 21 {i think it is obv on cases like xxx0111..1xxxx} use ur brain. // subtask 5 {more than half score} // as long as 2^(10-x) > 1000/(2^x) we can find it. xxxxxxx0111111 //