Serval soon said goodbye to Japari kindergarten, and began his life in Japari Primary School.ios
In his favorite math class, the teacher taught him the following interesting definitions.算法
A parenthesis sequence is a string, containing only characters "(" and ")".express
A correct parenthesis sequence is a parenthesis sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, parenthesis sequences "()()", "(())" are correct (the resulting expressions are: "(1+1)+(1+1)", "((1+1)+1)"), while ")(" and ")" are not. Note that the empty string is a correct parenthesis sequence by definition.ide
We define that |s||s| as the length of string ss. A strict prefix s[1…l]s[1…l] (1≤l<|s|)(1≤l<|s|) of a string s=s1s2…s|s|s=s1s2…s|s| is string s1s2…sls1s2…sl. Note that the empty string and the whole string are not strict prefixes of any string by the definition.学习
Having learned these definitions, he comes up with a new problem. He writes down a string ss containing only characters "(", ")" and "?". And what he is going to do, is to replace each of the "?" in ss independently by one of "(" and ")" to make all strict prefixes of the new sequence not a correct parenthesis sequence, while the new sequence should be a correct parenthesis sequence.this
After all, he is just a primary school student so this problem is too hard for him to solve. As his best friend, can you help him to replace the question marks? If there are many solutions, any of them is acceptable.atom
The first line contains a single integer |s||s| (1≤|s|≤3⋅1051≤|s|≤3⋅105), the length of the string.spa
The second line contains a string ss, containing only "(", ")" and "?".rest
A single line contains a string representing the answer.code
If there are many solutions, any of them is acceptable.
If there is no answer, print a single line containing ":(" (without the quotes).
6
(?????
(()())
10
(???(???(?
:(
It can be proved that there is no solution for the second sample, so print ":(".
题目大意:给定一个只含有'(', ')', '?'三种字符的字符串,其中'?'能够转换成'(' 或 ')',要求判断这个字符串是否符合除首尾任何'('的前面都没有')'且整个字符串为合法序列。
题解:这里学习了xyq学长的假设算法
#include<iostream> #include<string>
using namespace std; const int N = 300005; int n, pos[N];//pos后缀意味着逆序时这个点有多少个')'
string s; int main() { ios::sync_with_stdio(false); cin >> n; cin >> s; if (n % 2) { cout << ":(" << endl; return 0; } if (s[0] == ')' || s[n - 1] == '(') { cout << ":(" << endl; return 0; } if (s[0] == '?') { s[0] = '('; } if (s[n - 1] == '?') { s[n - 1] = ')'; } for (int i = n - 2; i >= 1; i--) { pos[i] = pos[i + 1]; if (s[i] == ')' || s[i] == '?')//开局假设全部的 '?' 都是都是')'
pos[i]++; else pos[i]--; } int lsum = 0; for (int i = 1; i < n - 1; i++) { if (s[i] == '(') lsum++; else if (s[i] == ')') { if (lsum == 0) { cout << ":(" << endl; return 0; } else lsum--; } else { if (lsum) { if (pos[i] > lsum) { s[i] = '('; lsum++; } else { s[i] = ')'; lsum--; } } else { s[i] = '('; lsum++; } } } if (lsum) { cout << ":(" << endl; } else cout << s << endl; return 0; }