leetcode

class :
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        ans = []
        for query in queries:
            val, ind = query
            A[ind] += val
            temp = 0
            for a in A:
                if (a % 2 == 0):
                    temp += a
            ans.append(temp)
        return ans

# 改进后（better，but still TLE）
class :
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        ans = []
        even_id = []
        temp = 0
        for ind, a in enumerate(A):
            if (a % 2 == 0):
                even_id.append(ind)
                temp += a
        for query in queries:
            val, ind = query
            A[ind] += val
            if (ind in even_id and A[ind] % 2 == 0):
                temp += val
            elif (ind in even_id and A[ind] % 2 != 0):
                temp -= (A[ind] - val)
                even_id.remove(ind)
            elif (ind not in even_id and A[ind] % 2 == 0):
                temp += A[ind]
                even_id.append(ind)
            else:
                pass
            ans.append(temp)
        return ans

# 官方解法
class :
    def sumEvenAfterQueries(self, A: List[int], queries: List[List[int]]) -> List[int]:
        S = sum(x for x in A if x % 2 == 0) # 这里和我想法一致
        ans = []

        for x, k in queries:
            if A[k] % 2 == 0: S -= A[k]
            A[k] += x
            if A[k] % 2 == 0: S += A[k] # 这里我考虑了所有情况，但其实没有必要，因此TLE
            ans.append(S)

        return ans