题目如下：

解题思路：就三个字-线段树。这个题目是线段树用法最经典的场景。

代码如下：

class NumArray(object):    def __init__(self, nums):        """        :type nums: List[int]        """        self.nl = nums        self.tree = []        if len(nums) == 0:            return        for i in xrange((4*len(nums)+1)):            self.tree.append([])        self.build(1,len(nums),1,nums)        #print self.tree    def build(self,l,r,k,nums):        self.tree[k] = [l,r]        if l == r: #leaf            self.tree[k].append(nums[l-1])            return nums[l-1]        wl = self.build(l,(l+r)/2,2*k,nums)        wr = self.build((l+r)/2+1,r,2*k+1,nums)        #print l,r,wl,wr        self.tree[k].append(wl+wr)        return self.tree[k][2]    def update(self, i, val):        """        :type i: int        :type val: int        :rtype: void        """        if i > len(self.nl):            return        diff = self.nl[i] - val        #for j in range(i,len(self.sl)):        #    self.sl[j] -= diff        self.nl[i] = val        k = 1        i += 1        while True:            self.tree[k][2] -= diff            m = (self.tree[k][0] + self.tree[k][1])/2            if self.tree[k][0] == self.tree[k][1]:                break            if i <= m:                k = 2*k            else:                k = 2*k + 1        #print self.tree    def calcRange(self,i,j,k):        #print i,j,k        if i == self.tree[k][0] and j == self.tree[k][1]:            return self.tree[k][2]        m = (self.tree[k][0] + self.tree[k][1])/2        if j <= m:            return self.calcRange(i,j,2*k)        elif i > m:            return self.calcRange(i,j,2*k+1)        else:            return self.calcRange(i,m,2*k) + self.calcRange(m+1,j,2*k+1)    def sumRange(self, i, j):        """        :type i: int        :type j: int        :rtype: int        """        #print self.sl        #print self.nl        return self.calcRange(i+1,j+1,1)        # Your NumArray object will be instantiated and called as such:# obj = NumArray(nums)# obj.update(i,val)# param_2 = obj.sumRange(i,j)