(1) 对于静态数组多次区间查询, 采用前缀和处理以优化时间;

(2) 题目要等差求和, 于是初始化等差前缀和数组ladder;

(3) 对于ladder[r] - ladder[l - 1], 区间内每个元素都多加了l - 1倍,

于是减去(l - 1) * (sum[r] - sum[l - 1]),

对此需要初始化sum前缀和数组.

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;

const int mod = 998244353;
const int N = 1e5 + 5;

int n, q, l, r;
int a[N], ladder[N], sum[N];

signed main()
{
    cin >> n >> q;
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        ladder[i] = (ladder[i - 1] + (i * a[i]) % mod) % mod;
        sum[i] = (sum[i - 1] + a[i]) % mod;
    }

    while(q--) {
        cin >> l >> r;
        int x = (ladder[r] - ladder[l - 1] + mod) % mod;
        int y = (l - 1) * ((sum[r] - sum[l - 1] + mod) % mod) % mod;
        cout << (x - y + mod) % mod << endl;
    }
}