ACM思維題訓練集合
A bracket sequence is a string containing only characters “(” and “)”.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters “1” and “+” between the original characters of the sequence. For example, bracket sequences “()()”, “(())” are regular (the resulting expressions are: “(1)+(1)”, “((1+1)+1)”), and “)(” and “(” are not.

You are given n bracket sequences s1,s2,…,sn. Calculate the number of pairs i,j(1≤i,j≤n) such that the bracket sequence si+sj is a regular bracket sequence. Operation + means concatenation i.e. “()(” + “)()” = “()()()”.

If si+sj and sj+si are regular bracket sequences and i≠j, then both pairs (i,j) and (j,i) must be counted in the answer. Also, if si+si is a regular bracket sequence, the pair (i,i) must be counted in the answer.

Input
The first line contains one integer n(1≤n≤3?105) — the number of bracket sequences. The following n lines contain bracket sequences — non-empty strings consisting only of characters “(” and “)”. The sum of lengths of all bracket sequences does not exceed 3?105.

Output
In the single line print a single integer — the number of pairs i,j(1≤i,j≤n) such that the bracket sequence si+sj is a regular bracket sequence.

Examples
Input
3
)
()
(
Output
2
Input
2
()
()
Output
4
Note
In the first example, suitable pairs are (3,1) and (2,2).

In the second example, any pair is suitable, namely (1,1),(1,2),(2,1),(2,2).
模擬稍微有一下就可以了

#include <bits/stdc++.h> using namespace std; template <typename t> void read(t &x) {char ch = getchar();x = 0;t f = 1;while (ch < '0' || ch > '9')f = (ch == '-' ? -1 : f), ch = getchar();while (ch >= '0' && ch <= '9')x = x * 10 + ch - '0', ch = getchar();x *= f; }#define wi(n) printf("%d ", n) #define wl(n) printf("%lld ", n) #define rep(m, n, i) for (int i = m; i < n; ++i) #define rrep(m, n, i) for (int i = m; i > n; --i) #define P puts(" ") typedef long long ll; #define MOD 1000000007 #define mp(a, b) make_pair(a, b) #define N 1005 #define fil(a, n) rep(0, n, i) read(a[i]) //---------------https://lunatic.blog.csdn.net/-------------------// map<LL, LL> mp; char con[N]; int main() {LL i, p, j, n, check;LL cont = 0, ans = 0, len1, len2;scanf("%lld", &n);getchar();for (j = 1; j <= n; j++){p = check = 0;len1 = len2 = 0;memset(con, 0, sizeof(0));scanf("%s", con);for (i = 0; i < 300009; i++){if (con[i] == 0)break;if (con[i] == '('){len1++;p++;}else{p--;if (len1)len1--;elselen2++;}}if (len1 == 0 && len2 == 0)cont++;else{if (len1 == 0)mp[p]++;if (len2 == 0)mp[p]++;}}ans = cont * cont;map<LL, LL>::iterator it1;for (it1 = mp.begin(); it1 != mp.end(); it1++){if (it1->first > 0)break;if (mp[-(it1->first)] > 0)ans += (it1->second) * mp[-(it1->first)];}printf("%lld\n", ans);return 0; } 創作挑戰賽新人創作獎勵來咯，堅持創作打卡瓜分現金大獎

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