题目见此

分析，把‘（’当成1， ‘）’当成-1， 计算前缀和sum。

记交换括号左边的序号为u， 右边为v，讨论左右括号：

1、s[u] == ‘(’ && s[v] == ‘)’ 那么[u, v - 1]的前缀和会全部-2

2、s[u] == ‘(’ && s[v] == ‘(’ 显然

3、s[u] == ‘)’ && s[v] == ‘(’ 那么[u, v - 1]的前缀和会全部+2

4、s[u] == ‘)’ && s[v] == ‘)’ 显然

对于一个括号序列，序列合法的要求是对∃sum[i]>=0 && sum[N]=0， 所以2，3，4的变换都不会改变序列的合法性。对于第一种情况，只要min(sum[i])>=2,i∈[u,v−1]即可。可以用ST或者线段树实现快速查询最小值。

/*****************************************************/

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <map>

#include <set>

#include <ctime>

#include <stack>

#include <queue>

#include <cmath>

#include <string>

#include <vector>

#include <cstdio>

#include <cctype>

#include <cstring>

#include <sstream>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)

#define   sigma_size    26

#define   lson          l,m,v<<1

#define   rson          m+1,r,v<<1|1

#define   slch          v<<1

#define   srch          v<<1|1

#define   sgetmid       int m = (l+r)>>1

#define   LL            long long

#define   ull           unsigned long long

#define   mem(x,v)      memset(x,v,sizeof(x))

#define   lowbit(x)     (x&-x)

#define   bits(a)       __builtin_popcount(a)

#define   mk            make_pair

#define   pb            push_back

#define   fi            first

#define   se            second

const int    INF    = 0x3f3f3f3f;

const LL     INFF   = 1e18;

const double pi     = acos(-1.0);

const double inf    = 1e18;

const double eps    = 1e-9;

const LL     mod    = 1e9+7;

const int    maxmat = 10;

const ull    BASE   = 31;

/*****************************************************/

const int maxn = 2e5 + 5;

int N, Q;

char s[maxn];

int a[maxn];

int seg[maxn << 2], col[maxn << 2];

void PushUp(int v) {

    seg[v] = min(seg[slch], seg[srch]);

}

void build(int l, int r, int v) {

    if (l == r) seg[v] = a[l];

    else {

        sgetmid;

        build(lson);

        build(rson);

        PushUp(v);

    }

}

int query(int L, int R, int l, int r, int v) {

    if (L <= l && r <= R) return seg[v];

    sgetmid;

    int ans = INF;

    if (L <= m) ans = min(ans, query(L, R, lson));

    if (R >  m) ans = min(ans, query(L, R, rson));

    return ans;

}

int main(int argc, char const *argv[]) {

    int N, Q;

    while (~scanf("%d%d", &N, &Q)) {

        mem(seg, 0); mem(a, 0);

        scanf("%s", s);

        for (int i = 1; i <= N; i ++) {

            if (s[i - 1] == '(') a[i] = a[i - 1] + 1;

            else a[i] = a[i - 1] - 1;

        }

        build(1, N, 1);

        for (int i = 0; i < Q; i ++) {

            int u, v;

            scanf("%d%d", &u, &v);

            if (u > v) swap(u, v);

            if (s[u - 1] == '(' && s[v - 1] == ')') {

                int pos = query(u, v - 1, 1, N, 1);

                if (pos >= 2) puts("Yes");

                else puts("No");

            }

            else puts("Yes");

        }

    }

    return 0;

}