快速排序

确定分界点：x = q[(l + r) / 2], l, r
调整范围左边大于x，右边小于x
递归处理左边和右边

#include <iostream>

using namespace std;

const int N =  100010;

int q[N];
int n;

void quick_sort(int q[], int l, int r){
    if (l >= r) return;
    int x = q[l + r >> 1], i = l - 1, j = r + 1;//先加一再比较
    while(i < j){
        do ++ i; while(q[i] < x);
        do -- j; while(q[j] > x);
        if(i < j) swap(q[i], q[j]);
    }

    quick_sort(q, l, j);
    quick_sort(q, j + 1, r);//左右两边分别递归
}

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

    for(int i = 0; i < n; i ++){
        scanf("%d", &q[i]); 
    }

    quick_sort(q, 0, n - 1);

    for(int i = 0; i < n; i ++)
        printf("%d ", q[i]);

    return 0;
}

应用:

第k个数

#include <iostream>

using namespace std;

const int N = 100010;

int n, k;
int q[N];

int quick_sort_k(int q[], int l, int r, int k){
    if( l >= r) return q[l];
    int x = q[l + r >> 1];
    int i = l - 1, j = r + 1;

    while(i < j){
        do i ++; while(q[i] < x);
        do j --; while(q[j] > x);

        if(i < j) swap(q[i], q[j]);
    }

    if(j - l + 1 >= k) return quick_sort_k(q, l, j, k);//如果k在l到j中
    else return quick_sort_k(q, j + 1, r, k - (j - l + 1));
}

int main()
{
    cin >> n >> k;

    for(int i = 0; i < n; i ++) cin >> q[i];

    cout << quick_sort_k(q, 0, n - 1, k) << endl;

    return 0;
}

归并排序

确定分界点mid = (l + r) / 2，[L, R] => [L, mid], [mid + 1, R]
递归排序[L, mid]和[mid + 1, R]
归并,双指针表示剩余部分中最小元素的位置,左右两个序列合并成一个有序序列

#include <iostream>

using namespace std;

const int N = 1e5 + 10;

int a[N], tmp[N];

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];
}

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);

    merge_sort(a, 0, n - 1);

    for (int i = 0; i < n; i ++ ) printf("%d ", a[i]);

    return 0;
}

应用：

逆序对的数量

此时q[i] > q[j], 则q[i]后面的数都大于q[j]

#include <iostream>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int a[N], tmp[N];

LL merge_sort(int q[], int l, int r)
{
    if (l >= r) return 0;

    int mid = l + r >> 1;

    LL res = 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
        {
            res += mid - i + 1;
            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];

    return res;
}

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);

    cout << merge_sort(a, 0, n - 1) << endl;

    return 0;
}

二分

模板

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

// 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;
}

check(mid) 检查mid是否满足某种性质
如果需要查找的边界是区间的右端点，满足该条件为true，此时l = mid，false为r = mid - 1（l + r + 1）
如果需要查找的边界是区间的左端点，满足该条件为true，此时r = mid，false为l = mid + 1

应用：

数的范围

#include <iostream>

using namespace std;

const int N = 100010;
int n, m;
int q[N];

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 0; i < n; i ++ ) scanf("%d", &q[i]);

    while(m --){
        int x;
        scanf("%d", &x);
        int l = 0, r = n - 1;
        while (l < r){
            int mid = l + r >> 1;
            if(q[mid] >= x) r = mid;
            else l = mid + 1;
        }

        if(q[l] != x) cout << "-1 -1"<< endl;
        else{
            cout << l <<' ';
            int l = 0, r = n - 1;
            while(l < r){
                int mid = (l + r + 1) >> 1;
                if(q[mid] <= x) l = mid;
                else r = mid - 1;
            }
            cout << l << endl;
        }
    }

    return 0;
}

浮点数二分

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;
}

应用

数的三次方根

#include <iostream>

using namespace std;

int main()
{
    double n;

    cin >> n;

    double l = -10000, r = 10000;
    while(r - l > 1e-8){
        double mid = (l + r) / 2;
        if(mid * mid * mid >= n) r = mid;
        else l = mid;
    }

    printf("%lf\n", l);

    return 0;
}

高精度计算

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;
    }

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

    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;
}