ZOJ-3279 Ants 树状数组 + 二分

Posted 2020-12-04 fzl194

题目链接：

https://cn.vjudge.net/problem/ZOJ-3279

题目大意：

有1到n 那个level 每一个level有a[i]只蚂蚁
两种操作 

p a b 把第a个level的蚂蚁数量改成b

q  a  查询第a只蚂蚁在哪个level里。

解题思路：

用树状数组动态维护前缀和，二分查找第a个蚂蚁即可

要用printf，不然会超时

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e5 + 10;
 4 int tree[maxn], n, a[maxn];
 5 int lowbit(int x)
 6 {
 7     return (x & (-x));
 8 }
 9 void add(int x, int d)//将x下标的值加上d
10 {
11     while(x <= n)
12     {
13         tree[x] += d;
14         x += lowbit(x);
15     }
16 }
17 int sum(int x)
18 {
19     int tot = 0;
20     while(x)
21     {
22         tot += tree[x];
23         x -= lowbit(x);
24     }
25     return tot;
26 }
27 
28 int main()
29 {
30     while(scanf("%d", &n) != EOF)
31     {
32         memset(a, 0, sizeof(a));
33         memset(tree, 0, sizeof(tree));
34         for(int i = 1; i <= n; i++)
35         {
36             scanf("%d", &a[i]);
37             add(i, a[i]);
38         }
39         int m, x, y;
40         scanf("%d", &m);
41         char s[2];
42         while(m--)
43         {
44             scanf("%s", s);
45             if(s[0] == ‘p‘)
46             {
47                 scanf("%d%d", &x, &y);
48                 add(x, -a[x]);//单点修改
49                 add(x, y);
50                 a[x] = y;
51             }
52             else if(s[0] == ‘q‘)
53             {
54                 scanf("%d", &x);
55                 int l = 1, r = n, ans;
56                 while(l <= r)
57                 {
58                     int m = (l + r) / 2;
59                     if(sum(m) >= x)ans = m, r = m - 1;
60                     else l = m + 1;
61                 }
62                 printf("%d
", ans);
63             }
64         }
65     }
66     return 0;
67 }