C++数论刷题记录 Hackerrank Solve Mathematics Constructing a Number / Closest Number

Constructing a Number 

Manipulating numbers is at the core of a programmer's job. To test how well you know their properties, you are asked to solve the following problem.

You are given  non-negative integers , , ..., . You want to know whether it's possible to construct a new integer using all the digits of these numbers such that it would be divisible by . You can reorder the digits as you want. The resulting number can contain leading zeros.

For example, consider the numbers  from which you have to construct a new integer as described above. Numerous arrangements of digits are possible; but we have illustrated one below.

注意：数据类型不能用int储存，否则tle（wa）

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);
vector<string> split(const string &);

/*
 * Complete the 'canConstruct' function below.
 *
 * The function is expected to return a STRING.
 * The function accepts INTEGER_ARRAY a as parameter.
 */

string canConstruct(vector<int> a) {
    // Return "Yes" or "No" denoting whether you can construct the required number.
    long long sum=0;
    for(int i=0;i<a.size();i++){
        sum+=a[i];
    }
    if(sum%3==0) return "Yes";
    else return "No";
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string t_temp;
    getline(cin, t_temp);

    int t = stoi(ltrim(rtrim(t_temp)));

    for (int t_itr = 0; t_itr < t; t_itr++) {
        string n_temp;
        getline(cin, n_temp);

        int n = stoi(ltrim(rtrim(n_temp)));

        string a_temp_temp;
        getline(cin, a_temp_temp);

        vector<string> a_temp = split(rtrim(a_temp_temp));

        vector<int> a(n);

        for (int i = 0; i < n; i++) {
            int a_item = stoi(a_temp[i]);

            a[i] = a_item;
        }

        string result = canConstruct(a);

        fout << result << "\n";
    }

    fout.close();

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

vector<string> split(const string &str) {
    vector<string> tokens;

    string::size_type start = 0;
    string::size_type end = 0;

    while ((end = str.find(" ", start)) != string::npos) {
        tokens.push_back(str.substr(start, end - start));

        start = end + 1;
    }

    tokens.push_back(str.substr(start));

    return tokens;
}

Closest Number

You are given 3 numbers a, b and x. You need to output the multiple of x which is closest to ab. If more than one answer exists , display the smallest one.

Input Format

The first line contains T, the number of testcases.
T lines follow, each line contains 3 space separated integers (a, b and x respectively)

Constraints

1 ≤ T ≤ 105
1 ≤ x ≤ 109
0 < ab ≤ 109
1 ≤ a ≤ 109
-109 ≤ b ≤ 109

Output Format

For each test case , output the multiple of x which is closest to ab

Sample Input 0

3
349 1 4
395 1 7
4 -2 2

Sample Output 0

348
392
0

Explanation 0

The closest multiple of 4 to 349 is 348.
The closest multiple of 7 to 395 is 392.
The closest multiple of 2 to 1/16 is 0.

注意：b=0的情况需要单独判断x的值，否则只能输出0.并且判断多个multiple的情况，接近a^b才是目标

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);
vector<string> split(const string &);

/*
 * Complete the 'closestNumber' function below.
 *
 * The function is expected to return an INTEGER.
 * The function accepts following parameters:
 *  1. INTEGER a
 *  2. INTEGER b
 *  3. INTEGER x
 */
//a=8 b=0 x=1
int closestNumber(int a, int b, int x) {
    if(b==0) {
        if(x==1) return x;
        else return 0;
    }
    long long power = pow(a,b);
    long long lower = (power/x)*x; //first multiple 
    long long next = lower+(power>0?x:-x); //second multiple
    if(abs(lower-power)<=abs(next-power)){
        return lower;
    }
    else {
    return next;
    }
}

int main()
{
    ofstream fout(getenv("OUTPUT_PATH"));

    string t_temp;
    getline(cin, t_temp);

    int t = stoi(ltrim(rtrim(t_temp)));

    for (int t_itr = 0; t_itr < t; t_itr++) {
        string first_multiple_input_temp;
        getline(cin, first_multiple_input_temp);

        vector<string> first_multiple_input = split(rtrim(first_multiple_input_temp));

        int a = stoi(first_multiple_input[0]);

        int b = stoi(first_multiple_input[1]);

        int x = stoi(first_multiple_input[2]);

        int result = closestNumber(a, b, x);

        fout << result << "\n";
    }

    fout.close();

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

vector<string> split(const string &str) {
    vector<string> tokens;

    string::size_type start = 0;
    string::size_type end = 0;

    while ((end = str.find(" ", start)) != string::npos) {
        tokens.push_back(str.substr(start, end - start));

        start = end + 1;
    }

    tokens.push_back(str.substr(start));

    return tokens;
}