【HDU1166】敌兵布阵（树状数组或线段树）

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cstdio>
 5 #include <numeric>
 6 #include <algorithm>
 7 #include <cmath>
 8 #include <cctype>
 9 #include <string>
10 using namespace std;
11 
12 
13 const int N = 50005;
14 int C[N];
15 
16 struct peo {
17     int x, v;
18 }P[N];
19 
20 int lowbit (int x) {
21     return x & -x;
22 }
23 
24 int sum (int x) {
25     int ret = 0;
26     while (x > 0) {
27         ret += C[x]; x -= lowbit(x);
28     }
29     return ret;
30 }
31 
32 void add (int x, int d) {
33     while (x <= 50000) {
34         C[x] += d; x += lowbit(x);
35     }
36 }
37 
38 int _sum (int d, int u) {
39     return sum(u) - sum(d - 1);
40 }
41 
42 int main () {
43     int T, n, cur = 0; scanf("%d", &T);
44     char op[10];
45     while (T --) {
46         memset(C, 0, sizeof(C));
47         memset(P, 0, sizeof(P));
48         scanf("%d", &n);
49         for (int i = 0 ; i < n; ++ i) {
50             scanf("%d", &P[i].v);
51             P[i].x = i + 1;
52             add (P[i].x, P[i].v);
53         }
54         //cout << sum(3) << endl;
55         printf("Case %d:\n", ++ cur);
56         while (scanf("%s", op)) {
57             if (op[0] == ‘E‘) {
58                 break;
59             } else if (op[0] == ‘Q‘) {
60                 int x, y;
61                 scanf("%d %d", &x, &y);
62                 cout << _sum(x, y) << endl;
63             } else if (op[0] == ‘S‘) {
64                 int x, y;
65                 scanf("%d %d", &x, &y);
66                 add (x, -y);
67             } else if (op[0] == ‘A‘) {
68                 int x, y;
69                 scanf("%d %d", &x, &y);
70                 add (x, y);
71             }
72         }
73     }
74     return 0;
75 }

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <vector>
 5 #include <cstring>
 6 #include <cstdlib>
 7 #include <cctype>
 8 #include <algorithm>
 9 #include <numeric>
10 
11 using namespace std;
12 
13 struct P{
14     int sum, l, r;
15 } node[150050];
16 
17 int n, num[50050] = {0};
18 
19 void build (int k, int l, int r) {
20     node[k].l = l, node[k].r = r;
21     if (l == r) {
22         node[k].sum = num[l];
23     } else {
24         build(2 * k, l, (l + r) / 2);
25         build(2 * k + 1, ((l + r) / 2) + 1, r);
26         node[k].sum = node[2 * k].sum + node[2 * k + 1].sum;
27     }
28 }
29 
30 void update(int k, int i, int x) {
31     if (node[k].l <= i && node[k].r >= i) {
32         node[k].sum += x;
33     }
34     if (node[k].l == node[k].r) {
35         return ;
36     }
37     if (i <= node[2 * k].r) {
38         update(2 * k, i, x);
39     }
40     else update(2 * k + 1, i, x);
41 }
42 
43 int query (int k, int l, int r) {
44     if (l == node[k].l && r == node[k].r) {
45         return node[k].sum;
46     }
47     if (r <= node[2 * k].r)  {
48         return query(2 * k, l, r);
49     }
50     if (l >= node[2 * k + 1].l) {
51         return query(2 * k + 1, l, r);
52     } else {
53         return query(2 * k, l, node[2 * k].r) + query(2 * k + 1, node[2 * k + 1].l, r);
54     }
55 }
56 
57 int main(){
58     int T;
59     scanf("%d",&T);
60     for (int z = 1; z <= T; ++ z) {
61         memset(num, 0, sizeof(num));
62         scanf("%d", &n);
63         for (int i = 1; i <= n; ++ i) {
64             scanf("%d",&num[i]);
65         }
66         build(1, 1, n);
67         char s[100];
68         printf("Case %d:\n",z);
69         while (scanf("%s", s)){
70             if (s[0] == ‘E‘)    break;
71             if (s[0] == ‘Q‘){
72                 int a, b;
73                 scanf("%d%d" ,&a ,&b);
74                 printf("%d\n",query(1, a, b));
75             }
76             if (s[0] == ‘A‘){
77                 int a,x;
78                 scanf("%d%d", &a, &x);
79                 update(1, a, x);
80             }
81             if (s[0] == ‘S‘) {
82                 int a,x;
83                 scanf("%d%d", &a, &x);
84                 update(1, a, -x);
85             }
86         }
87     }
88     return 0;
89 }