acwing算法第二课

一、高精度加法--A+B    10⁶

　　大整数从个位（小位）开始存在数组里

// C = A + B, A >= 0, B >= 0
vector<int> add(vector<int> &A, vector<int> &B)
{
    if (A.size() < B.size()) return add(B, A);

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

    if (t) C.push_back(t);
    return C;
}

作者：yxc
链接：https://www.acwing.com/blog/content/277/
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。

二、高精度减法

　　和加法差不多，但要先判断A是否大于B，要分情况讨论。

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;
        if (i < B.size()) t -= B[i];
        C.push_back((t + 10) % 10);
        if (t < 0) t = 1;
        else t = 0;
    }

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

作者：yxc
链接：https://www.acwing.com/blog/content/277/
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。

三、高精度乘法

　　高精度乘低精度:

// C = A * b, A >= 0, b >= 0
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();

    return C;
}

作者：yxc
链接：https://www.acwing.com/blog/content/277/
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。

四、高精度除法

　　高精度除以低精度：

 

 

// A / b = C ... r, A >= 0, b > 0
vector<int> div(vector<int> &A, int b, int &r)
{
    vector<int> C;
    r = 0;
    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;
}

作者：yxc
链接：https://www.acwing.com/blog/content/277/
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。

 

 如何将高精度整数以字符串写入，再读到数组中：

 

五、前缀和

　　一维：

S[i] = a[1] + a[2] + ... a[i]
a[l] + ... + a[r] = S[r] - S[l - 1]

　　二维：

S[i, j] = 第i行j列格子左上部分所有元素的和
以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵的和为：
S[x2, y2] - S[x1 - 1, y2] - S[x2, y1 - 1] + S[x1 - 1, y1 - 1]

六、差分

　　一维：

给区间[l, r]中的每个数加上c：B[l] += c, B[r + 1] -= c

#include <iostream>

using namespace std;

const int N = 100010;

int n, m;
int a[N], b[N];

void insert(int l, int r, int c)
{
    b[l] += c;
    b[r + 1] -= c;
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &a[i]);

    for (int i = 1; i <= n; i ++ ) insert(i, i, a[i]);

    while (m -- )
    {
        int l, r, c;
        scanf("%d%d%d", &l, &r, &c);
        insert(l, r, c);
    }

    for (int i = 1; i <= n; i ++ ) b[i] += b[i - 1];

    for (int i = 1; i <= n; i ++ ) printf("%d ", b[i]);

    return 0;
}

作者：yxc
链接：https://www.acwing.com/activity/content/code/content/39799/
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。

　　

　　二维：

#include <iostream>

using namespace std;

const int N = 1010;

int n, m, q;
int a[N][N], b[N][N];

void insert(int x1, int y1, int x2, int y2, int c)
{
    b[x1][y1] += c;
    b[x2 + 1][y1] -= c;
    b[x1][y2 + 1] -= c;
    b[x2 + 1][y2 + 1] += c;
}

int main()
{
    scanf("%d%d%d", &n, &m, &q);

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            scanf("%d", &a[i][j]);

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            insert(i, j, i, j, a[i][j]);

    while (q -- )
    {
        int x1, y1, x2, y2, c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        insert(x1, y1, x2, y2, c);
    }

    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= m; j ++ )
            b[i][j] += b[i - 1][j] + b[i][j - 1] - b[i - 1][j - 1];

    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 1; j <= m; j ++ ) printf("%d ", b[i][j]);
        puts("");
    }

    return 0;
}

作者：yxc
链接：https://www.acwing.com/activity/content/code/content/39800/
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。

浅浅记录一下y总的课，顺便搞个板子。