金山云-9.23 C++后台开发笔试

• 按照从上到下，从左往右顺序输入一颗二叉树中每一个结点对应的权重，判断该满二叉树是否存在这样的非叶子节点，使得左数所有叶子结点之和等于右树所有叶子结点之和，如果存在输出Yes，不存在输出No。

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <numeric>
#include <sstream>
using namespace std;

int sum(vector<int>v, int index)
{
    if (index<v.size())
        return v[index] + sum(v, 2 * index + 1) + sum(v, 2 * index + 2);
    else return 0;
}
int main()
{
    //input  demo
    vector<vector<int>> v;
    int count;
    cin >> count;
    getchar();
    for (int i = 0; i < count; i++)
    {
        vector<int> temp;
        string str;
        getline(cin, str);
        stringstream ss;
        ss << str;
        string number;
        while (getline(ss, number, ' '))
        {
            temp.push_back(stoi(number));
        }
        v.push_back(temp);
    }
    //
    for (int k = 0; k < v.size(); k++)
    {
        int flag = 0;
        for (int i = 0; i <= (v[k].size() / 2 - 1); i++)
        {
            if (sum(v[k], 2 * i + 1) == sum(v[k], 2 * i + 2))
            {
                cout << "Yes" << endl;
                flag = 1;
                break;
            }
        }
        if (flag == 0) cout << "No" << endl;
    }
}
           

• 输入正整数n，判断1-n之间整数出现5的次数（暴力解法）

#include <iostream>
using namespace std;

int findFive(int n){
    int res = 0;
    for (int i = 1; i <= n; i++) {
        int temp = i;
        while (temp > 0) {
            if (temp % 10 == 5) {
                res++;
            }
            temp = temp / 10;
        }
    }
    return res;
}

int main() {
    int n;
    cin >> n;
    cout << findFive(n) << endl;
    return 0;
}