Sereja and Brackets CodeForces - 380C （线段树+分治思路）

Sereja and Brackets

Sereja has a bracket sequence s1, s2, ..., s**n, 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 l**i, r**i(1 ≤ l**i ≤ r**i ≤ 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, ..., s**n (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 l**i, r**i (1 ≤ l**i ≤ r**i ≤ 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.

Examples

Input

())(())(())(71 12 31 21 128 125 112 10

Output

00210466

Note

A subsequence of length |x| of string s = s1s2... s|s| (where |s| is the length of string s) is string x = s**k1s**k2... s**k|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：

给你一个只含有'(' 和')' 的字符串，

以及q个询问，每一个询问给你两个整数l和r，代表一个区间。对于每一个询问，让你输出区间中能选出最长的子序列是合法的括号序列的长度。

思路：

线段树+分治的思想来解决此问题。

我们线段树每一个区间维护以下信息：

1、区间中能选出最长的子序列是合法的括号序列的个数 num。

2、区间中多余的'(' 字符的个数 a

3、区间中多余的')' 字符的个数 b

那么对于区间合并时，

num=左儿子的num+右儿子的num+min（左儿子的a，右儿子的b）

a=左儿子的a+右儿子的a - min（左儿子的a，右儿子的b）

b=左儿子的b+右儿子的b - min（左儿子的a，右儿子的b）

最后输出时，注意num个括号个数，*2才是长度。

细节见代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <queue>

#include <stack>

#include <map>

#include <set>

#include <vector>

#include <iomanip>

#define ALL(x) (x).begin(), (x).end()

#define sz(a) int(a.size())

#define all(a) a.begin(), a.end()

#define rep(i,x,n) for(int i=x;i<n;i++)

#define repd(i,x,n) for(int i=x;i<=n;i++)

#define pii pair<int,int>

#define pll pair<long long ,long long>

#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)

#define MS0(X) memset((X), 0, sizeof((X)))

#define MSC0(X) memset((X), '\0', sizeof((X)))

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define eps 1e-6

#define gg(x) getInt(&x)

#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl

using namespace std;

typedef long long ll;

ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}

ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}

ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2) { ans = ans * a % MOD; } a = a * a % MOD; b /= 2;} return ans;}

inline void getInt(int *p);

const int maxn = 1000010;

const int inf = 0x3f3f3f3f;

/*** TEMPLATE CODE * * STARTS HERE ***/

struct node {

    int l, r;

    int num;

    int a;// (

    int b;// )

} segmeng_tree[maxn << 2];

char s[maxn];

int n;

int m;

void pushup(int rt)

{

    int x = min(segmeng_tree[rt << 1].a, segmeng_tree[rt << 1 | 1].b);

    segmeng_tree[rt].num = x + segmeng_tree[rt << 1].num + segmeng_tree[rt << 1 | 1].num;

    segmeng_tree[rt].a = segmeng_tree[rt << 1].a + segmeng_tree[rt << 1 | 1].a - x;

    segmeng_tree[rt].b = segmeng_tree[rt << 1].b + segmeng_tree[rt << 1 | 1].b - x;

}

void build(int rt, int l, int r)

{

    segmeng_tree[rt].l = l;

    segmeng_tree[rt].r = r;

    if (l == r) {

        segmeng_tree[rt].a = s[l] == '(';

        segmeng_tree[rt].b = s[l] == ')';

        segmeng_tree[rt].num = 0;

    } else {

        int mid = (l + r) >> 1;

        build(rt << 1, l, mid);

        build(rt << 1 | 1, mid + 1, r);

        pushup(rt);

    }

}

node ask(int rt, int l, int r)

{

    if (segmeng_tree[rt].l >= l && segmeng_tree[rt].r <= r) {

        return segmeng_tree[rt];

    }

    int mid = (segmeng_tree[rt].l + segmeng_tree[rt].r) >> 1;

    if (r <= mid) {

        return ask(rt << 1, l, r);

    } else if (l > mid) {

        return ask(rt << 1 | 1, l, r);

    } else {

        node res1 = ask(rt << 1, l, r);

        node res2 = ask(rt << 1 | 1, l, r);

        node res = res1;

        int x = min(res1.a, res2.b);

        res.num += x;

        res.b += res2.b;

        res.a += res2.a;

        res.num += res2.num;

        res.b -= x;

        res.a -= x;

        return res;

    }

}

int main()

{

    //freopen("D:\\code\\text\\input.txt","r",stdin);

    //freopen("D:\\code\\text\\output.txt","w",stdout);

    scanf("%s", s + 1);

    n = strlen(s + 1);

    build(1, 1, n);

    scanf("%d", &m);

    while (m--) {

        int l, r;

        scanf("%d %d", &l, &r);

        printf("%d\n", ask(1, l, r).num * 2);

    }

    return 0;

}

inline void getInt(int *p)

{

    char ch;

    do {

        ch = getchar();

    } while (ch == ' ' || ch == '\n');

    if (ch == '-') {

        *p = -(getchar() - '0');

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 - ch + '0';

        }

    } else {

        *p = ch - '0';

        while ((ch = getchar()) >= '0' && ch <= '9') {

            *p = *p * 10 + ch - '0';

        }

    }

}