1 /*
2 线段树-成段更新：裸题，成段增减，区间求和
3 注意：开long long:)
4 */
5 #include <cstdio>
6 #include <iostream>
7 #include <algorithm>
8 #include <cstring>
9 #include <cmath>
10 using namespace std;
11
12 #define lson l, mid, rt << 1
13 #define rson mid + 1, r, rt << 1 | 1
14 #define LL long long
15
16 const int MAXN = 1e5 + 10;
17 const int INF = 0x3f3f3f3f;
18 struct Node
19 {
20 LL sum, add;
21 }node[MAXN << 2];
22
23 void push_up(int rt)
24 {
25 node[rt].sum = node[rt<<1].sum + node[rt<<1|1].sum;
26 }
27
28 void push_down(int rt, int c)
29 {
30 if (node[rt].add)
31 {
32 node[rt<<1].add += node[rt].add;
33 node[rt<<1|1].add += node[rt].add;
34 node[rt<<1].sum += node[rt].add * (c - (c >> 1));
35 node[rt<<1|1].sum += node[rt].add * (c >> 1);
36 node[rt].add = 0;
37 }
38 }
39
40 void build(int l, int r, int rt)
41 {
42 node[rt].add = 0;
43 if (l == r) {scanf ("%I64d", &node[rt].sum); return ;}
44 int mid = (l + r) >> 1;
45 build (lson); build (rson);
46
47 push_up (rt);
48 }
49
50 void updata(int ql, int qr, int c, int l, int r, int rt)
51 {
52 if (ql <= l && r <= qr) {node[rt].sum += (LL) c * (r - l + 1); node[rt].add += c; return ;}
53
54 push_down (rt, r - l + 1);
55
56 int mid = (l + r) >> 1;
57 if (ql <= mid) updata (ql, qr, c, lson);
58 if (qr > mid) updata (ql, qr, c, rson);
59
60 push_up (rt);
61 }
62
63 LL query(int ql, int qr, int l, int r, int rt)
64 {
65 if (ql <= l && r <= qr) return node[rt].sum;
66
67 push_down (rt, r - l + 1);
68
69 int mid = (l + r) >> 1; LL ans = 0;
70 if (ql <= mid) ans += query (ql, qr, lson);
71 if (qr > mid) ans += query (ql, qr, rson);
72
73 return ans;
74 }
75
76 int main(void) //POJ 3468 A Simple Problem with Integers
77 {
78 //freopen ("POJ_3468.in", "r", stdin);
79
80 int n, q;
81 while (scanf ("%d%d", &n, &q) == 2)
82 {
83 build (1, n, 1);
84 while (q--)
85 {
86 char s[3]; int ql, qr, c;
87 scanf ("%s", &s);
88 if (s[0] == 'Q')
89 {
90 scanf ("%d%d", &ql, &qr);
91 printf ("%I64d
", query (ql, qr, 1, n, 1));
92 }
93 else
94 {
95 scanf ("%d%d%d", &ql, &qr, &c);
96 updata (ql, qr, c, 1, n, 1);
97 }
98 }
99 }
100
101 return 0;
102 }