F题，求解啊！！！

#include<iostream>
#include<set>
using namespace std;
typedef long long LL;
const int N = 10000010,MOD = 998244353;
LL n,m,a[N],p[N];
set<int> pos,zr;

LL go(int st,int l,int r)
{
    LL ans = (l == st ? 1ll : 2ll) * a[st],pos = st;
    while(pos + 1 <= r && pos - st <= 30)
    {
        pos ++;
        ans = 2ll * max(ans,a[pos]);
    }
    ans %= MOD;
    ans = ans * p[r - pos] % MOD;
    return (ans % MOD + MOD) % MOD;
}

int main()
{
    p[0] = 1;
    for(int i = 1;i < N;i++) p[i] = 2ll * p[i - 1] % MOD;
    scanf("%lld%lld",&n,&m);
    for(int i = 1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        if(a[i] > 0) pos.insert(i);
        if(a[i] == 0) zr.insert(i);
    }
    int op;
    LL x,v,l,r;
    while(m -- )
    {
        scanf("%d",&op);
        if(op == 1)
        {
            scanf("%lld%lld",&x,&v);
            a[x] += v;
            if(a[x] > 0) pos.insert(x);
            if(a[x] == 0) zr.insert(x);
            if(a[x] <= 0 && pos.count(x)) pos.erase(x);
            if(a[x] != 0 && zr.count(x)) zr.erase(x);
        }
        else
        {
            scanf("%lld%lld",&l,&r);
            int p = n + 1;
            if(pos.lower_bound(l) != pos.end()) p = *pos.lower_bound(l);
            if(p > r)
            {
                int z = n + 1;
                if(zr.lower_bound(l) != zr.end()) z = *zr.lower_bound(l);
                if(z <= r) puts("0");
                else printf("%lld\n",go(max(l,r - 30),l,r));
            }
            else printf("%lld\n",go(p,l,r));
        }
    }
    return 0;
}

第11行为什么需要做一个判断 l == st ? 1ll : 2ll啊，我不理解，这样为什么就不会爆long long 了，有没有佬解答一下