RUST MLE 经典考题。论 IO 的实现

这里用ffi的io或者其他办法的io是有办法规避io引起的MLE的，但是思路一样，无所谓啦

fn solve(input: &mut FastReader<impl BufRead>, output: &mut BufWriter<impl Write>) -> Result<()> {
    _ = input.nextln();
    let mut arr = input.readln::<i64>();
    while input.has_next() {
        let tmp = input.readln::<usize>();
        if tmp[0] == 1 {
            let [_, l, r, k] = &tmp[..] else { todo!() };
            arr[*l - 1..*r].rotate_right(k % (r - l + 1));
        } else {
            let [_, i] = &tmp[..] else { todo!() };
            writeln!(output, "{}", arr[i - 1])?;
        }
    }
    Ok(())
}

#include<iostream>
#include<vector>
using namespace std;
typedef long long ll;
const int N = 1e7 + 10;
int n, q;
vector<ll>a(N);
void roll(vector<ll>& a, int l, int r)
{
    for (int i = l, j = r; i <j; i++, j--)
    {
       swap(a[i],a[j]);
    }
}
int main()
{
    scanf("%d%d",&n,&q);
    for (int i = 0; i < n; i++)
    {
        scanf("%lld",&a[i]);
    }
    while (q--)
    {
        int op;
        scanf("%d",&op);
        if (op == 1)
        {
            int l, r, k;
            scanf("%d%d%d",&l,&r,&k);
            l--, r--;
            int len = r - l + 1;
            k %= len;
            roll(a, l, r), roll(a, l, l + k - 1), roll(a, l + k, r);
        }
        else
        {
            int x;
            scanf("%d",&x);
            printf("%lld\n",a[x-1]);
        }
    }
    return 0;
}