/*

 * @lc app=leetcode.cn id=371 lang=c

 *

 * [371] 两整数之和

 *

 * https://leetcode-cn.com/problems/sum-of-two-integers/description/

 *

 * algorithms

 * Easy (55.04%)

 * Total Accepted:    8.1K

 * Total Submissions: 14.8K

 * Testcase Example:  '1\n2'

 *

 * 不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。

 *

 * 示例 1:

 *

 * 输入: a = 1, b = 2

 * 输出: 3

 *

 *

 * 示例 2:

 *

 * 输入: a = -2, b = 3

 * 输出: 1

 *

 */

int getSum(int a, int b) {

    if(a && b) return getSum(a^b, (a&b) << );

    else return a|b;

}

#

# @lc app=leetcode.cn id=371 lang=python3

#

# [371] 两整数之和

#

# https://leetcode-cn.com/problems/sum-of-two-integers/description/

#

# algorithms

# Easy (55.04%)

# Total Accepted:    8.1K

# Total Submissions: 14.8K

# Testcase Example:  '1\n2'

#

# 不使用运算符 + 和 - ​​​​​​​，计算两整数 ​​​​​​​a 、b ​​​​​​​之和。

#

# 示例 1:

#

# 输入: a = 1, b = 2

# 输出: 3

#

#

# 示例 2:

#

# 输入: a = -2, b = 3

# 输出: 1

#

#

class Solution:

    def getSum(self, a: int, b: int) -> int:

        while b != 0:

            carry = a & b

            a = (a ^ b) % 0x100000000

            b = (carry << 1) % 0x100000000

        return a if a <= 0x7FFFFFFF else a | (~0x100000000+1)