思路：

線段樹區間更新。

實現：

 1 #include <cstdio>
 2 #include <cstring>
 3 using namespace std;
 4 typedef long long ll;
 5 const int N = 100005;
 6 ll tree[N * 4], lazy[N * 4], a[N];
 7 int n, m;
 8 
 9 void build(int num, int l, int r)
10 {
11     if (l == r) { tree[num] = a[l]; return; }
12     int m = l + r >> 1
 
;
13     build(num * 2, l, m);
14     build(num * 2 + 1, m + 1, r);
15     tree[num] = tree[num * 2] + tree[num * 2 + 1];
16 }
17 
18 void pushdown(int num, int cl, int cr)
19 {
20     if (!lazy[num]) return;
21     tree[num * 2] += lazy[num] * cl;
22     tree[num * 2 + 1] += lazy[num] * cr;
23     lazy[num * 2
 
] += lazy[num];
24     lazy[num * 2 + 1] += lazy[num];
25     lazy[num] = 0;
26 }
27 
28 void update(int num, int l, int r, int x, int y, ll dx)
29 {
30     if (x <= l && y >= r) { tree[num] += (r - l + 1) * dx; lazy[num] += dx; return; }
31     int m = l + r >> 1;
32     pushdown(num, m - l + 1
 
, r - m);
33     if (x <= m) update(num * 2, l, m, x, y, dx);
34     if (y >= m + 1) update(num * 2 + 1, m + 1, r, x, y, dx);
35     tree[num] = tree[num * 2] + tree[num * 2 + 1];
36 }
37 
38 ll query(int num, int l, int r, int x, int y)
39 {
40     if (x <= l && y >= r) return tree[num];
41     int m = l + r >> 1;
42     pushdown(num, m - l + 1, r - m);
43     ll ans = 0;
44     if (x <= m) ans += query(num * 2, l, m, x, y);
45     if (y >= m + 1) ans += query(num * 2 + 1, m + 1, r, x, y);
46     return ans;
47 }
48 
49 int main()
50 {
51     scanf("%d %d", &n, &m);
52     for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
53     getchar();
54     build(1, 1, n);
55     char tmp; int x, y; ll d;
56     for (int i = 0; i < m; i++)
57     {
58         scanf("%c", &tmp);
59         if (tmp == ‘Q‘) { scanf("%d %d", &x, &y); getchar(); printf("%lld\n", query(1, 1, n, x, y)); }
60         else { scanf("%d %d %lld", &x, &y, &d); getchar(); update(1, 1, n, x, y, d); }
61     }
62     return 0;
63 }

poj3468 A Simple Problem with Integers