E1. Cats on the Upgrade (easy version)

// Problem: E1. Cats on the Upgrade (easy version)
// Contest: Codeforces - Codeforces Round #765 (Div. 2)
// URL: https://codeforces.com/contest/1625/problem/E1
// Memory Limit: 256 MB
// Time Limit: 6000 ms
// 2022-03-09 00:22:15
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long
void solve() {
    int n, m;
    cin >> n >> m;
    string s;
    cin >> s;
    s = " " + s;
    stack<int> st; 
    vector<int> f(n + 5, 0), sum(n + 5, 0);
    for (int i = 1; i <= n; i ++) {
        char ch = s[i];
        if (ch == '(') {
            st.push(i );
        }
        else {
            if (st.empty()) continue;
            f[i] = f[st.top() - 1] + 1;
            st.pop();
        }
        sum[i] = sum[i - 1] + f[i];
    }
    while( m--) {
        int l, r;
        int t;
        cin >> t >> l >> r;
        int ans = sum[r]  -sum[l - 1] - f[l - 1] * (f[r] - f[l - 1]);
        printf("%lld\n",ans);
    }
}
signed main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

f[i]表示出现在i前面的左右括号的数量 f[i] = f[s.top() - 1] + 1; 然后对于答案 sum[r] - sum[l - 1] - f[l - 1] * (f[r] - f[l - 1])后面是去重 对于错的公式也有价值。像这题通过去重就可以得到正确的答案