Codeforces 380C. Sereja and Brackets（模拟啊）

题目链接：http://codeforces.com/problemset/problem/380/C

Sereja has a bracket sequence s1, s2, ..., sn, or, in other words, a string s of length n, consisting of characters "(" and ")".

Sereja needs to answer m queries, each of them is described by two integers li, ri (1 ≤ li ≤ ri ≤ n). The answer to the i-th query is the length of the maximum correct bracket subsequence of sequence sli, sli + 1, ..., sri. Help Sereja answer all queries.

You can find the definitions for a subsequence and a correct bracket sequence in the notes.

Input

The first line contains a sequence of characters s1, s2, ..., sn (1 ≤ n ≤ 106) without any spaces. Each character is either a "(" or a ")". The second line contains integer m (1 ≤ m ≤ 105) — the number of queries. Each of the next m lines contains a pair of integers. The i-th line contains integers li, ri (1 ≤ li ≤ ri ≤ n) — the description of the i-th query.

Output

Print the answer to each question on a single line. Print the answers in the order they go in the input.

Sample test(s)

input

())(())(())(
7
1 1
2 3
1 2
1 12
8 12
5 11
2 10

output

0
0
2
10
4
6
6

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is string x = sk1sk2... sk|x|(1 ≤ k1 < k2 < ... < k|x| ≤ |s|).

A correct bracket sequence is a bracket sequence that can be transformed into a correct aryphmetic expression by inserting characters "1" and "+" between the characters of the string. For example, bracket sequences "()()", "(())" are correct (the resulting expressions "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

For the third query required sequence will be «()».

For the fourth query required sequence will be «()(())(())».

题意：

给出一个含‘ ( ’ ，‘ ) ' 的字符串，

然后给出一个询问区间，去这个区间里完整匹配的（）的数目！一个（）的答案为2！

PS:

我们可以用栈把每个’ （ ‘的下标存起来！然后再扫一遍’ ）‘！用树状数组维护一下即可！

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <stack>
using namespace std;
#define maxn 1000017
#define maxm 100017
using namespace std;

struct node
{
    int l, r;
    int id;
} q[maxm];

char s[maxn];
int c[maxn];
int ans[maxm];
stack<int> st;
int len;

int Lowbit(int x)  // 2^k
{
    return x&(-x);
}

void update(int i, int x)//i点增量为x
{
    while(i <= len)
    {
        c[i] += x;
        i += Lowbit(i);
    }
}

int sum(int x)//区间求和 [1,x]
{
    int ans = 0;
    while(x)
    {
        ans += c[x];
        x -= Lowbit(x);
    }
    return ans;
}

bool cmp(node a, node b)
{
    return a.r < b.r;
}

int main()
{
    int m;

    while(~scanf("%s",s))
    {
        int pos = 1;
        memset(c, 0, sizeof(c));
        while(!st.empty())
        {
            st.pop();
        }
        len = strlen(s);
        scanf("%d",&m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d",&q[i].l,&q[i].r);
            q[i].id = i;
        }
        sort(q, q+m, cmp);
        for(int i = 0; i < m; i++)
        {
            for(int j = pos; j <= q[i].r; j++)
            {
                if(s[j-1] == '(')
                {
                    st.push(j);
                }
                else if(s[j-1] == ')')
                {
                    if(!st.empty())
                    {
                        int tt = st.top();
                        update(tt, 2);
                        st.pop();
                    }
                }
            }
            pos = q[i].r+1;
            ans[q[i].id] = sum(q[i].r) - sum(q[i].l-1);
        }
        for (int i = 0; i < m; i++)
        {
            printf("%d\n", ans[i]);
        }
    }
    return 0;
}

/*
())(())(())(
7
1 1
1 2
2 3
2 10
5 11
1 12
8 12
*/