Codeforces 380 C. Sereja and Brackets

树状数组求区间内范围个数。。。。


C. Sereja and Brackets
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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

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|).

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 ?()?.



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=1001000;

int tr1[maxn],tr2[maxn],tot;
char str[maxn];

int lowbit(int x)
{
    return x&(-x);
}

void Add(int tr[],int n)
{
    while(n<maxn)
    {
        tr[n]++;
        n+=lowbit(n);
    }
}

int Sum(int tr[],int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=tr[x];
        x-=lowbit(x);
    }
    return sum;
}

struct Interval
{
    int l,r,len,ans,id;
}q[maxn],a[maxn];

bool cmp1(Interval a,Interval b)
{
    if(a.len!=b.len)
    {
        return a.len<b.len;
    }
    else  return a.l<b.l;
}

bool cmpID(Interval a,Interval b)
{
    return a.id<b.id;
}

int st[maxn],pop=-1,m,qt;

int main()
{
    scanf("%s",&str);
    int len=strlen(str);
    for(int i=0;i<len;i++)
    {
        if(str[i]==‘(‘)
        {
            st[++pop]=i+1;
        }
        else
        {
            if(pop>=0)
            {
                a[tot].l=st[pop--];
                a[tot].r=i+1;
                a[tot].len=a[tot].r-a[tot].l+1;
                tot++;
            }
        }
    }

    scanf("%d",&m);
    for(int i=0;i<m;i++)
    {
        int l,r;
        scanf("%d%d",&l,&r);
        q[qt].l=l,q[qt].r=r;
        q[qt].len=r-l+1,q[qt].id=i;
        qt++;
    }
    sort(q,q+qt,cmp1);
    sort(a,a+tot,cmp1);

    int pos=0;
    for(int i=0;i<m;i++)
    {
        while(pos<tot&&a[pos].len<=q[i].len)
        {
            Add(tr1,a[pos].l);Add(tr2,a[pos].r);
            pos++;
        }

        q[i].ans=Sum(tr2,q[i].r)-Sum(tr1,q[i].l-1);
    }

    sort(q,q+qt,cmpID);

    for(int i=0;i<qt;i++)
    {
        printf("%d\n",q[i].ans*2);
    }

    return 0;
}



Codeforces 380 C. Sereja and Brackets

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