题目地址：https://leetcode-cn.com/problems/sum-of-digits-in-the-minimum-number/
题目描述
Given an array A of positive integers, let S be the sum of the digits of the minimal element of A.
Return 0 if S is odd, otherwise return 1.
Example 1:
Input: [34,23,1,24,75,33,54,8]
Output: 0
Explanation:
The minimal element is 1, and the sum of those digits is S = 1 which is odd, so the answer is 0.
Example 2:
Input: [99,77,33,66,55]
Output: 1
Explanation:
The minimal element is 33, and the sum of those digits is S = 3 + 3 = 6 which is even, so the answer is 1.
Note:
1、 1<=A.length<=100；
2、 1<=A[i].length<=100；
题目大意
给你一个正整数的数组 A。 然后计算 S，使其等于数组 A 当中最小的那个元素各个数位上数字之和。 最后，假如 S 所得计算结果是 奇数 的请你返回 0，否则请返回 1。
解题方法
遍历
先找出最小的数字，然后求其各位数字的和。
C++代码如下：
class Solution {
public:
int sumOfDigits(vector<int>& A) {
int min_num = INT_MAX;
for (int a : A) {
min_num = min(min_num, a);
}
int k = 0;
int s = 0;
while (min_num != 0) {
s += min_num % 10;
k++;
min_num /= 10;
}
return 1 - (s & 1);
}
};
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