5-17-template.md

第一讲

经典排序

快速排序算法模板

void quick_sort(int q[], int l, int r) {
    if (l >= r) return;
    
    int i = l - 1, j = r + 1, x = q[l];
    while (i < j) {
        do i ++ ; while (q[i] < x);
        do j -- ; while (q[j] > x);
        if (i < j) swap(q[i], q[j]);
        else break;
    }
    quick_sort(q, l, j), quick_sort(q, j + 1, r);
}

归并排序算法模板

void merge_sort(int q[], int l, int r) {
    if (l >= r) return;
    
    int mid = l + r >> 1;
    merge_sort(q, l, mid);
    merge_sort(q, mid + 1, r);
    
    int k = 0, i = l, j = mid + 1;
    while (i <= mid && j <= r)
        if (q[i] < q[j]) tmp[k ++ ] = q[i ++ ];
        else tmp[k ++ ] = q[j ++ ];
    
    while (i <= mid) tmp[k ++ ] = q[i ++ ];
    while (j <= r) tmp[k ++ ] = q[j ++ ];
    
    for (i = l, j = 0; i <= r; i ++, j ++ ) q[i] = tmp[j];
}

二分

整数二分算法模板

// 检查x是否满足某种性质
bool check(int x) {
    /* ... */
}

// 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用：
int bsearch_1(int l, int r) {
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid)) r = mid;    // check()判断mid是否满足性质
        else l = mid + 1;
    }
    return l;
}

// 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
int bsearch_2(int l, int r) {
    while (l < r) {
        int mid = l + r + 1 >> 1;
        if (check(mid)) l = mid;
        else r = mid - 1;
    }
    return l;
}

浮点数二分算法模板

bool check(double x) {/* ... */} // 检查x是否满足某种性质

double bsearch_3(double l, double r)
{
    const double eps = 1e-6;   // eps 表示精度，取决于题目对精度的要求
    while (r - l > eps)
    {
        double mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }
    return l;
}

高精度问题

高精度加法

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;
}

高精度减法

C = A - B, 满足A >= B, A >= 0, B >= 0

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;
}

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;
    }
    
    return C;
}

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;
}

前缀和

一维前缀和

S[i] = a[1] + a[2] + ... a[i]

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

二维前缀和

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]

差分

一维差分

B[i] = a[i] - a[i - 1]

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

二维差分

给以(x1, y1)为左上角，(x2, y2)为右下角的子矩阵中的所有元素加上c：

S[x1, y1] += c, S[x2 + 1, y1] -= c, S[x1, y2 + 1] -= c, S[x2 + 1, y2 + 1] += c