tree array
Below a represents the original array, and c represents the tree array
树状数组功能：
1.给某个位置上的数加上一个数
2.快速的求前缀和
c[x]层数的确定：
看x的二进制表示末尾有几个0，有几个0就是第几层
c[x]含义表示：
lowbit(x) = 2^k (k 是x二进制表示末尾0的个数)
c[x] 表示在A数组中一段区间的和 ：c[x] = ( x - 2^k, x] = ( x - lowbit(x), x]
1. Add a number to the number at a certain position
// 当前点是x，父节点是x + lowbit(x)，若给x加上v，父节点都要同时加上v
a[x] += v;
for (int i = x; i <= n; i += lowbit(i)) c[i] += v;
2. Quickly find the prefix sum
int res = 0;
for (int i = x; i > 0; i -= lowbit(i)) res += c[i];
Example: Find the sum of continuous intervals dynamically
Given a sequence of n numbers, two operations are specified, one is to modify a certain element, and the other is to obtain the continuous sum of the sub-sequence [a,b].
input format
The first line contains two integers n and m, which respectively represent the number of numbers and the number of operations.
The second line contains n integers, representing the complete sequence.
Next m lines, each line contains three integers k, a, b (k=0, means finding the sum of the subsequence [a,b]; k=1, means adding b to the ath number).
Sequences are counted from 1.
output format
Output several rows of numbers, which represent the continuous sum of the corresponding sub-sequences [a,b] when k=0.
data range
1 ≤ n ≤ 100000,
1 ≤ m ≤ 100000,
1 ≤ a ≤ b ≤ n,
data guarantee that at any time, the sum of all elements in the array is within the range of int.
Input sample:
10 5
1 2 3 4 5 6 7 8 9 10
1 1 5
0 1 3
0 4 8
1 7 5
0 4 8
Sample output:
11
30
35
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 100010;
int a[N], c[N];
int n, m;
int lowbit(int x)
{
return x & -x;
}
// 在x的位置上加上v
void add(int x, int v)
{
for (int i = x; i <= n; i += lowbit(i)) c[i] += v;
}
// 查询1 - x位置的和
int query(int x)
{
int res = 0;
for (int i = x; i > 0; i -= lowbit(i)) res += c[i];
return res;
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++) scanf("%d", &a[i]);
for (int i = 1; i <= n; i ++) add(i, a[i]);
while (m --)
{
int k, x, y;
scanf("%d%d%d", &k, &x, &y);
if (k == 0) printf("%d\n", query(y) - query(x - 1));
else add(x, y);
}
return 0;
}