线段树满江红

#include <cstdio>
using namespace std;
#define int long long
const int N = 200005;
int n, m, a[N], val[N << 2], tag[N << 2];
void build(int p, int sta, int end) {
	if (sta == end) {
		val[p] = a[sta];
		return;
	}
	int mid = (sta + end) >> 1;
	build(2 * p, sta, mid);
	build(2 * p + 1, mid + 1, end);
	val[p] = val[p * 2] + val[p * 2 + 1];
}
void push_down(int p, int sta, int end) {
	if (tag[p] && sta != end) {
		int mid = (sta + end) >> 1;
		val[2 * p] += tag[p] * (mid - sta + 1), val[2 * p + 1] += tag[p] * (end - mid);
		tag[2 * p] += tag[p], tag[2 * p + 1] += tag[p];
		tag[p] = 0;
	}
}
int ask(int p, int sta, int end, int l, int r) {
	if (l <= sta && end <= r)
		return val[p];
	push_down(p, sta, end);
	int mid = (sta + end) >> 1, sum = 0;
	if (l <= mid) sum += ask(2 * p, sta, mid, l, r);
	if (r > mid) sum += ask(2 * p + 1, mid + 1, end, l, r);
	return sum;
}
void add(int p, int sta, int end, int l, int r, int k)  {
	if (l <= sta && end <= r) {
		val[p] += k * (sta - end + 1), tag[p] += k;
		return;
	}
	push_down(p, sta, end);
	int mid = (sta + end) >> 1;
	if (l <= mid) add(2 * p, sta, mid, l, r, k);
	if (r > mid) add(2 * p + 1, mid + 1, end, l, r, k);
	val[p] = val[2 * p] + val[2 * p + 1];
}
signed main() {
	scanf("%lld%lld", &n, &m);
	for (int i = 1; i <= n; i ++)
		scanf("%lld", &a[i]);
	build(1, 1, n);
	for (int i = 1; i <= m; i ++) {
		int opt, l, r, k;
		scanf("%lld", &opt);
		if (opt == 1) {
			scanf("%lld%lld%lld", &l, &r, &k);
			add(1, 1, n, l, r, k);
		} else if (opt == 2) {
			scanf("%lld", &k);
			add(1, 1, n, 1, 1, k);
		} else if (opt == 3) {
			scanf("%lld", &k);
			add(1, 1, n, 1, 1, -k);
		} else if (opt == 4) {
			scanf("%lld%lld", &l, &r);
			printf("%lld\n", ask(1, 1, n, l, r));
		} else if (opt == 5) {
			printf("%lld\n", ask(1, 1, n, 1, 1));
		}
	}
	return 0;
}