定义S(n)，表示n在十进制下的各位数字和。

现在给定一个x,请你求出最小正整数n，满足x<=S(n)。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
#include <iostream>
#include<vector>
#include<algorithm>
#include<string>

int main()
{
    int num;
    cin >> num;
    vector<string> resV;
    string a = "9";
    for (int n = 0; n < num; n++)
    {
        int x;
        cin >> x;
        //long long int multi = 1;
        string res;
        while (x > 9)
        {
            x -= 9;
            res = a + res;
            //multi *= 10;
        }
        if (x != 0)
        {
            //res += multi * x;
            res = to_string(x) + res;
        }
        resV.push_back(res);
    }
    for (int n = 0; n < num; n++)
    {
        cout << resV[n] << endl;
    }
    return 0;
}

小易给定你数字A, B (A < B)和系数p, q。每次操作你可以将A变成A + p或者将p变成p * q。问至少几次操作使得B <= A。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
#include <iostream>
#include<vector>
#include<algorithm>
#include<string>

int main()
{
    int num;
    cin >> num;
    vector<long long int> resV;
    
    for (int n = 0; n < num; n++)
    {
        long long int a;
        long long int b;
        long long int p;
        long long int q;
        cin >> a >> b >> p >> q;
        long long int res = 0;
        if (b - a > p)
        {
            long long int temp = b - a;
            long long int multiNum = p;

            while (temp > multiNum)
            {
                multiNum *= q;
                res++;
            }
            res++;
        }
        else
        {
            res++;
        }
        resV.push_back(res);
    }
    for (int n = 0; n < num; n++)
    {
        cout << resV[n] << endl;
    }
    return 0;
}

小易定义一个数字序列是完美的，当且仅当对于任意2 <= i <= n，都满足img，即每个数字都要大于等于前面所有数字的和。
现在给定数字序列Ai，小易想请你从中找出最长的一段连续子序列，满足它是完美的。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
#include <iostream>
#include<vector>
#include<algorithm>
#include<string>


int main()
{
    vector<int> resV;
    int num;
    cin >> num;
    for (int i = 0; i < num; i++)
    {
        int tempLength = 1;
        int maxLength = 0;
        vector<int> n;
        int serialNum;
        cin >> serialNum;
        for (int j = 0; j < serialNum; j++)
        {
            int data;
            cin >> data;
            n.push_back(data);
        }
        int compute = n[0];
        for (int k = 1; k < serialNum; k++)
        {
            if (n[k] >= compute)
            {
                compute += n[k];
                tempLength++;
            }
            else
            {
                maxLength = max(maxLength, tempLength);
                tempLength = 1;
                compute = n[k];
            }
        }
        resV.push_back(maxLength);
    }
    for (int i = 0; i < num; i++)
    {
        cout << resV[i] << endl;
    }
    return 0;
}

小易的公司一共有n名员工, 第i个人每个月的薪酬是xi万元。
现在小易的老板向小易提了m次询问, 每次询问老板都会给出一个整数k, 小易要快速回答老板工资等于k的员工的数量。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
#include <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<unordered_map>

using namespace std;
int main()
{

    int people, question;
    cin >> people >> question;
    vector<int> resV(question, 0);
    vector<int> salaryV;
    vector<int> questionV;
    unordered_map<string, size_t> questionMap;
    for (int i = 0; i < people; i++)
    {
        int salary;
        cin >> salary;
        salaryV.push_back(salary);
    }
    for (int i = 0; i < question; i++)
    {
        int data;
        cin >> data;
        questionV.push_back(data);
        questionMap[to_string(data)];
    }
    for (int i = 0; i < people; i++)
    {
        if (questionMap.find(to_string(salaryV[i])) != questionMap.end())
        {
            ++questionMap[to_string(salaryV[i])];
        }
    }

   
    for (int i = 0; i < question; i++)
    {
        cout << questionMap.find(to_string(questionV[i]))->second << endl;
    }
    return 0;
}