算法基础课

Count # of 1 in binary

https://www.acwing.com/problem/content/803/

Code

#include <iostream>

using namespace std;
inline int lowbit(int x) {
    return x & -x;
}
int main() {
    int n;
    cin >> n;
    while (n --) {
        int x, res = 0;
        cin >> x;

        while (x) {
            x -= lowbit(x);
            res++;
        }
        cout << res << ' ';
    }
}

a - b

https://www.acwing.com/problem/content/794/

Code

#include <iostream>
#include <vector>

using namespace std;

bool cmp(vector<int> &A, vector<int> &B)
{
    if (A.size() != B.size()) return A.size() > B.size(); // 大小不一致，长的大

    for (int i = A.size() - 1; i >= 0; i -- )
        if (A[i] != B[i]) // 左数第一个不一样的大小决定数字大小
            return A[i] > B[i];

    return true;
}

vector<int> sub(vector<int> &A, vector<int> &B)
{
    vector<int> C;
    for (int i = 0, t = 0; i < A.size(); i ++ )
    {
        t = A[i] - t; // 当前数减去借位的数
        if (i < B.size()) t -= B[i]; // 如果B数组比A数组短，那么剩余值减去B当前位
        C.push_back((t + 10) % 10); // 如果t < 0，取t + 10的值（0-6 = -4），同时借位，如果t >= 0，取%10，（5-3=2=2+10%10=2）
        if (t < 0) t = 1; // 借位
        else t = 0; // 不必借位
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a, b;
    vector<int> A, B;
    cin >> a >> b;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');
    for (int i = b.size() - 1; i >= 0; i -- ) B.push_back(b[i] - '0');

    vector<int> C;

    if (cmp(A, B)) C = sub(A, B);
    else C = sub(B, A), cout << '-';

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];
    cout << endl;

    return 0;
}

作者：yxc
链接：https://www.acwing.com/activity/content/code/content/39793/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

a * b

https://www.acwing.com/activity/content/code/content/39794/

Code

#include <iostream>
#include <vector>

using namespace std;


vector<int> mul(vector<int> &A, int b)
{
    vector<int> C;

    int t = 0;
    for (int i = 0; i < A.size() || t; i ++ )
    {
        if (i < A.size()) t += A[i] * b;
        C.push_back(t % 10);
        t /= 10;
    }

    while (C.size() > 1 && C.back() == 0) C.pop_back(); // 去除leading zeros

    return C;
}


int main()
{
    string a;
    int b;

    cin >> a >> b;

    vector<int> A;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');

    auto C = mul(A, b);

    for (int i = C.size() - 1; i >= 0; i -- ) printf("%d", C[i]);

    return 0;
}

作者：yxc
链接：https://www.acwing.com/activity/content/code/content/39794/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

a / b

https://www.acwing.com/activity/content/code/content/39795/

Code

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

vector<int> div(vector<int> &A, int b, int &r)
{
    vector<int> C;
    r = 0; // reminder
    for (int i = A.size() - 1; i >= 0; i -- )
    {
        r = r * 10 + A[i];
        C.push_back(r / b);
        r %= b;
    }
    reverse(C.begin(), C.end());
    while (C.size() > 1 && C.back() == 0) C.pop_back();
    return C;
}

int main()
{
    string a;
    vector<int> A;

    int B;
    cin >> a >> B;
    for (int i = a.size() - 1; i >= 0; i -- ) A.push_back(a[i] - '0');

    int r;
    auto C = div(A, B, r);

    for (int i = C.size() - 1; i >= 0; i -- ) cout << C[i];

    cout << endl << r << endl;

    return 0;
}

作者：yxc
链接：https://www.acwing.com/activity/content/code/content/39795/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

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Last update: August 17, 2020