First write code:
The first function
int LowBit(int x) //求 最小幂2^k
{ return x & (-x); } 第二个函数
int getSum(int pos) //求前n项和
{
int sum = 0;
while(pos > 0)
{
sum += arr[pos];
pos -= LowBit(pos);
}
return sum;
} The third function
void update(int pos , int val) //对某个元素进行加法操作：
{
while(pos <= n)
{
arr[pos] += val;
pos += LowBit(pos);
}
} http://blog.csdn.net/qq_34374664/article/details/52787481 explain Fenwick tree collection
Problem solving process, we sometimes need to maintain a prefix array and S [n-] = A [. 1] + A [2] + ... + A [n-] .
Difficult to find, however, if we modify any of the A [i], S [i ], S [i + 1] ... S [n] will change.
It can be said, each modified A [i], and to adjust the prefix S [] in the worst case will require O (n) time.
When n is very large, the program will run very slowly.
So, here we introduce the "Fenwick tree", it changes the sum is O (logn) , the efficiency is very high.
As shown, an array of red rectangle represents the C [] is an array we maintain tree .
Characteristics of the image: to form a C [8] is the root node, other C [i], A [j] is the sub-tree nodes.
For array A [] index is odd, which are just only a C [] elements, i.e., C [1] = A [1], C [3] = A [3], C [5] = A [ 5] etc.
For array A [] index is an even number, which is a series of A [i] and.
C [1] = A [1]; jurisdiction C [1] is 1
C [2] = A [2] + A [1]; jurisdiction C [2] is 2
C [3] = A [3]; jurisdiction C [3] is 1
C [4] = A [4] + A [3] + A [2] + A [1]; jurisdiction C [4] is 4
C [5] = A [5]; jurisdiction C [5] is 1
C [6] = A [6] + A [5]; jurisdiction C [6] is 2
C [7] = A [7]; jurisdiction C [7] is 1
C[8] = A[8] + A[7] + A[6] + A[5] + A[4] + A[3] + A[2] + A[1] ;C[8]的管辖范围为8
这里的管辖范围就是说C[]由几个A[]组成。
由上可得(S[]由C[]来表示)：
S[1] = C[1]
S[2] = C[2]
S[3] = C[3] + C[2]
S[4] = C[4]
S[5] = C[5] + C[4]
S[6] = C[6] + C[4]
S[7] = C[7] + C[6] + C[4]
S[8] = C[8]
通过观察图像：S[i]表示为i之前所有最大子树的根结点C[]累加的和
上面所说的管辖范围就是下面列出来的2^k所表示的数
| i | i所对应的二进制 |
2^k(k-->i的二进制的末尾0的个数) |
| 1 | 01 | 1 |
| 2 | 10 | 2 |
| 3 | 11 | 1 |
| 4 | 100 | 4 |
| 5 | 101 | 1 |
| 6 | 110 | 2 |
| 7 | 111 | 1 |
| 8 | 1000 | 8 |
2^k的求法：i 的二进制去掉所有高位的1，只保留最低位的1，如果只存在一个最高位的1，则 2^k 就是它本身。
如1（1），2（10），4（100），8（1000），16（10000）。即都是2的幂。
所以当我们判断一个数x是不是2的次幂时，只要判断x是否等于2^k就可以了，而2^k由位运算很容易求得（x & (-x)）。
对于C[i]如何求得他的父节点？
由上表格推断，C[i]的父节点为C[i+2^k]，也就是说将 i 的二进制的最末尾的1去掉，相邻的高位+1
对于C[i]如何求得他的下一个节点？
由上表格推断，C[i]的下一个节点是C[i-2^k]，也就是说将i的二进制最末尾的1去掉。
由上：
C[i]表示A[i-2^k+1]到A[i]的和，而k则是i在二进制时末尾0的个数。利用位运算，我们可以直接计算出2 ^ k = i & (i ^ (i - 1)) = (i & (-i))，所以C[i]的父节点为C[i + i & (-i)]。
这个k(i的二进制表示末尾0的个数)就是该节点在树中的高度，因而这个树的高度不会超过logn。所以,当我们修改A[i]的值时，可以从C[i]往根节点一路上溯，调整这条路上的所有C[]即可，这个操作的复杂度在最坏情况下就是树的高度即O(logn)。
而对于求数列的前n项和S[n]，只需找到C[n]以前（包括C[n]）的所有最大子树，把其根节点的C[]加起来即可。
不难发现，这些子树的数目是n在二进制表示时1的个数。因此，求和操作的复杂度也是O(logn)。http://blog.csdn.net/DGY01/article/details/52298912
http://blog.csdn.net/x_iya/article/details/8943264
http://blog.csdn.net/int64ago/article/details/7429868
