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2 years ago
#include <iostream>
using namespace std;
typedef long long LL;
#define ls u << 1
#define rs u << 1 | 1
const int N = 100010;
//快读
int read() {
int x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9') {
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
return x * f;
}
int n, q;
struct Node {
int l, r;
LL sum, tag; //区间总和,修改的数值(懒标记)
} tr[N << 2];
//更新统计信息
void pushup(int u) {
tr[u].sum = tr[ls].sum + tr[rs].sum;
}
//父节点向子孙节点传递信息
void pushdown(int u) {
if (tr[u].tag) {
//更新儿子的tag信息是为了更下一层的数据传递
// tag传递到子段节段的sum和需要按 区间长度*root.tag 进行增加
tr[ls].tag += tr[u].tag;
tr[ls].sum += LL(tr[ls].r - tr[ls].l + 1) * tr[u].tag;
tr[rs].tag += tr[u].tag;
tr[rs].sum += LL(tr[rs].r - tr[rs].l + 1) * tr[u].tag;
//别忘了清除懒标记
tr[u].tag = 0;
}
}
//构建
void build(int u, int l, int r) {
tr[u].l = l, tr[u].r = r; //标记范围
if (l == r) { //叶子
tr[u].sum = read(); //区间内只有一个元素l(r),区间和为read(),不需要记录向下的传递tag
return;
}
int mid = (l + r) >> 1;
build(ls, l, mid), build(rs, mid + 1, r); //左右儿子构建
pushup(u); //通过左右儿子构建后,向祖先节点反馈统计信息变化
}
//以u为根在区间[l,r]之间全都增加d
void add(int u, int l, int r, int d) {
if (tr[u].l >= l && tr[u].r <= r) {
tr[u].sum += LL(tr[u].r - tr[u].l + 1) * d; //总和增加
tr[u].tag += d; //懒标记+d
return;
}
pushdown(u); // 如果自己身上有旧的tag数值在递归前需要将原tag值pushdown到子孙节点去
int mid = tr[u].l + tr[u].r >> 1;
if (l <= mid) add(ls, l, r, d); //与左区间有交集
if (r > mid) add(rs, l, r, d); //与右区间有交集
pushup(u); //将结果的变更更新到祖先节点
}
//关键的查询操作
LL query(int u, int l, int r) {
if (tr[u].l >= l && tr[u].r <= r) return tr[u].sum;
/*懒标记的思想:在更新时只标识不修改,有查询时现用现改的思想。这么做的目的当然是怕有多组修改,
,3
*/
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
LL res = 0;
if (l <= mid) res = query(ls, l, r);
if (r > mid) res += query(rs, l, r);
//左查+ 右查 = 总和
return res;
}
int main() {
//文件输入输出
#ifndef ONLINE_JUDGE
freopen("P3372.in", "r", stdin);
#endif
n = read(), q = read();
//构建线段树
build(1, 1, n);
while (q--) {
int a, b, c, d;
a = read(), b = read(), c = read();
if (b > c) swap(b, c);
if (a == 1)
d = read(), add(1, b, c, d);
else
printf("%lld\n", query(1, b, c));
}
return 0;
}