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为了解决这个问题，我们需要处理基站报告的变化，并回答关于矩形区域内活动手机总数的查询。我们可以使用二维树状数组来高效地处理点更新和区间查询操作。

方法思路

解决代码

import sys
def main():
    # 读取输入直到遇到指令3
    while True:
        line = sys.stdin.readline()
        if not line:
            break
        line = line.strip()
        if not line:
            continue
        parts = list(map(int, line.split()))
        op = parts[0]
        if op == 3:
            break
        
        if op == 0:
            S = parts[1]
            S_val = S
            c = [[0]*(S_val +1) for _ in range(S_val +1)]
        elif op == 1:
            x = parts[1]
            y = parts[2]
            a = parts[3]
            # 更新坐标为x+1, y+1
            i = x +1
            j = y +1
            while i <= S_val:
                bit = i & -i
                while j <= S_val:
                    bit2 = j & -j
                    c[i][j] += a
                    j += bit2
                i += bit
        elif op == 2:
            l = parts[1]
            b = parts[2]
            r = parts[3]
            t = parts[4]
            # 转换为1-based
            l1 = l +1
            b1 = b +1
            r1 = r +1
            t1 = t +1
            # 计算区域和
            # sum(r1, t1) - sum(l1-1, t1) - sum(r1, b1-1) + sum(l1-1, b1-1)
            def get_sum(x, y):
                res = 0
                i = x
                while i > 0:
                    bit = i & -i
                    j = y
                    while j > 0:
                        res += c[i][j]
                        j -= bit
                    i -= bit
                return res
            
            total = get_sum(r1, t1) - get_sum(l1-1, t1) - get_sum(r1, b1-1) + get_sum(l1-1, b1-1)
            print(total)
        else:
            continue
if __name__ == "__main__":
    main()

代码解释

该方法能够高效处理多个指令，确保在合理时间内完成查询和更新操作。

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