• 方法 2：维护前缀和，$O(m \times n)$
• TLE，70 分

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

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	
	int n, m, cmd, x, k, l, r;
	cin >> n >> m;
	vector<int> a(n + 1);
	for (int i = 1; i <= n; i++) cin >> a[i];
	
	// f[i] = sum(a[1] + a[2] + ... + a[i])
	vector<int> f(n + 1, 0);
	for (int i = 1; i <= n; i++) f[i] = f[i - 1] + a[i];
	
	// O(m * n)
	while (m--) {
		cin >> cmd;
		if (cmd == 1) {
			cin >> x >> k;
			for (int i = x; i <= n; i++) f[i] += k;  // O(n)
		} else {
			cin >> l >> r;
			cout << f[r] - f[l - 1] << '\n';  // O(1)
		}
	}
	
	return 0;
}