POJ 3468

区间加减、查询,有了区间修改的经验,这题还是比较顺利的。

 1 #include <cstdio>
 2 using namespace std;
 3 
 4 #define recursive_def    int l, int r, int i
 5 #define lsi            i<<1
 6 #define rsi            lsi | 1
 7 #define lso            l, m, lsi
 8 #define rso            m+1, r, rsi
 9 #define pushup        (v[i] = v[lsi] + v[rsi])
10 #define pushdown    if (dirty[i]) {
11     dirty[lsi] += dirty[i], dirty[rsi] += dirty[i]; 
12     v[lsi] += dirty[i] * (m-l+1);    
13     v[rsi] += dirty[i] * (r-m);        
14     dirty[i] = 0;}
15 
16 #define MAXV    131072
17 
18 long long _v[MAXV<<1], *const v = &_v[-1];
19 long long _dirty[MAXV << 1], *const dirty = &_dirty[-1];
20 
21 void build(recursive_def)
22 {
23     dirty[i] = 0;
24     if (l == r) scanf("%lld", &v[i]);
25     else {
26         int m = l+r >> 1;
27         build(lso), build(rso);
28         pushup;
29     }
30 }
31 
32 int update(int L, int R, int val, recursive_def)
33 {
34     if (L<=l && r<=R) v[i] += (r-l+1) * val, dirty[i] += val;
35     else {
36         int m = l+r >> 1;
37         pushdown;
38         if (L <= m) update(L, R, val, lso);
39         if (m < R) update(L, R, val, rso);
40         pushup;
41     }
42 }
43 
44 long long query(int L, int R, recursive_def)
45 {
46     if (L<=l && r<=R) return v[i];
47     int m = l+r >> 1;
48     pushdown;
49     long long result = 0;
50     if (L <= m) result += query(L, R, lso);
51     if (m < R) result += query(L, R, rso);
52     return result;
53 }
54 
55 int main(void)
56 {
57 //    freopen("poj3468.txt", "r", stdin);
58     int N, Q;
59     while(scanf("%d%d", &N, &Q) > 0) {
60         build(1, N, 1);
61         while(Q) {
62             int L, R;
63             char op[3];
64             scanf("
%s", op);
65             if (op[0] == 'Q') {
66                 scanf("%d%d", &L, &R);
67                 printf("%lld
", query(L, R, 1, N, 1));
68             } else {
69                 int C;
70                 scanf("%d%d%d", &L, &R, &C);
71                 update(L, R, C, 1, N, 1);
72             }
73             --Q;
74         }
75     }
76     return 0;
77 }
3468 Accepted 4464K 2641MS G++ 1625B 2014-06-08 17:04:33
原文地址:https://www.cnblogs.com/e0e1e/p/poj_3468.html