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	`1`	`+If we meet a (`
	`2`	`+`
	`3`	`+Then () is a regular balanced string`
	`4`	`+`
	`5`	`+And (( is a palindrome`
	`6`	`+`
	`7`	`+No matter what, (x is always a good string and we will extract it`
	`8`	`+`
	`9`	`+-----`
	`10`	`+`
	`11`	`+If we meet a )`
	`12`	`+`
	`13`	`+Then, we cannot get a balanced string since it starts with )`
	`14`	`+`
	`15`	`+We can only get a palindrome`
	`16`	`+`
	`17`	`+All we need to do is find the next closed bracket !`
	`18`	`+`
	`19`	`+) ((((( ... ( )`
	`20`	`+`
	`21`	`+is a palindrome !`
	`22`	`+`
	`23`	`+-----`
	`24`	`+`
	`25`	`+void solve()`
	`26`	`+{`
	`27`	`+ int length;`
	`28`	`+ string S;`
	`29`	`+ cin >> length >> S;`
	`30`	`+`
	`31`	`+ int minimum_moves = 0, remaining = length;`
	`32`	`+ for(int i = 0; i + 1 < length; i++)`
	`33`	`+ {`
	`34`	`+ if(S[i] == '(')`
	`35`	`+ {`
	`36`	`+ minimum_moves++;`
	`37`	`+ remaining -= 2;`
	`38`	`+ i++;`
	`39`	`+ }`
	`40`	`+ else if(S[i] == ')')`
	`41`	`+ {`
	`42`	`+ int j = i + 1;`
	`43`	`+ while(j < length && S[j] != ')')`
	`44`	`+ {`
	`45`	`+ j++;`
	`46`	`+ }`
	`47`	`+`
	`48`	`+ if(j == length)`
	`49`	`+ {`
	`50`	`+ break;`
	`51`	`+ }`
	`52`	`+`
	`53`	`+ minimum_moves++;`
	`54`	`+ remaining -= (j - i + 1); //cout << "i = " << i << " j = " << j << "\n";`
	`55`	`+ i = j;`
	`56`	`+ }`
	`57`	`+ }`
	`58`	`+`
	`59`	`+ cout << minimum_moves << " " << remaining << "\n";`
	`60`	`+}`

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	`1`	`+We can always reach (x, y) in at most 2 moves`
	`2`	`+`
	`3`	`+If we are at the origin, then we need 0 moves`
	`4`	`+If (x^2 + y^2) is a perfect square, then we need only 1 move`
	`5`	`+Otherwise, we will go from (0, 0) -> (x, 0) -> (y, 0) in 2 moves !`
	`6`	`+`
	`7`	`+We will always need at most 2 moves !`
	`8`	`+`
	`9`	`+------`
	`10`	`+`
	`11`	`+int is_square(int n)`
	`12`	`+{`
	`13`	`+ for(int i = 1; i*i <= n; i++)`
	`14`	`+ {`
	`15`	`+ if(i*i == n)`
	`16`	`+ {`
	`17`	`+ return true;`
	`18`	`+ }`
	`19`	`+ }`
	`20`	`+`
	`21`	`+ return false;`
	`22`	`+}`
	`23`	`+`
	`24`	`+void solve()`
	`25`	`+{`
	`26`	`+ int x, y;`
	`27`	`+ cin >> x >> y;`
	`28`	`+`
	`29`	`+ int answer = 0;`
	`30`	`+ if(x == 0 && y == 0)`
	`31`	`+ {`
	`32`	`+ answer = 0;`
	`33`	`+ }`
	`34`	`+ else if(is_square(xx + yy))`
	`35`	`+ {`
	`36`	`+ answer = 1;`
	`37`	`+ }`
	`38`	`+ else`
	`39`	`+ {`
	`40`	`+ answer = 2;`
	`41`	`+ }`
	`42`	`+`
	`43`	`+ cout << answer << "\n";`
	`44`	`+}`

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	`1`	`+Be greedy and Set A[i] = A[i] + X, unless it exceeds the limit`
	`2`	`+`
	`3`	`+------`
	`4`	`+`
	`5`	`+void solve()`
	`6`	`+{`
	`7`	`+ int no_of_elements, x, y, max_limit;`
	`8`	`+ cin >> no_of_elements >> max_limit >> x >> y;`
	`9`	`+`
	`10`	`+ long long sum = 0;`
	`11`	`+ vector <long long> A(no_of_elements + 1);`
	`12`	`+ for(int i = 1; i <= no_of_elements; i++)`
	`13`	`+ {`
	`14`	`+ if(A[i - 1] + x <= max_limit)`
	`15`	`+ {`
	`16`	`+ A[i] = A[i - 1] + x;`
	`17`	`+ }`
	`18`	`+ else`
	`19`	`+ {`
	`20`	`+ A[i] = A[i - 1] - y;`
	`21`	`+ }`
	`22`	`+`
	`23`	`+ sum += A[i];`
	`24`	`+ }`
	`25`	`+`
	`26`	`+ cout << sum << "\n";`
	`27`	`+}`

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