1.快速排序

1.1快速排序

#include<stdio.h>
int n;
int q[1000];

void quick_sort(int q[],int l,int r)
//将数组q,数组首尾下标l,r输入快排函数quick_sort


{
    if(l>=r)return;
    //如果lr交叉，则说明数列已经被分割成单元素数列，排序完毕

    int x=q[(l+r)/2];
    //将数组中间位数值作为分界基准x
    int i=l-1;
    //设置从数组首位开始计数的指针i
    int j=r+1;
    //设置从数组末尾开始计数的指针j

    while(i<j)
    //在指针交叉前遍历一遍数组,将小于x的值置于左侧,大于x的值置于右侧
    {
        do i++;while(q[i]<x);
        //当i指向的数值小于分界基准x时继续向后移动
        //如数值大于x,则扣留当前元素，跳转到j指针
        do j--;while(q[j]>x);
        //当j指向的数值大于分界基准x时继续向前移动
        //当数值小于x,则将其与之前扣留的元素互换

        if(i<j)
        //如果指针未交叉,则执行交换
        {
            int temp=q[i];
            q[i]=q[j];
            q[j]=temp;
        }
        //交换完成后，继续让i开始移动
    }

    //递归,将x区分开的两部分分别再次进行处理
    quick_sort(q,l,j);
    quick_sort(q,j+1,r);
}
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;
}

1.2第k个数

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

using namespace std;

int main() {
    int n, k;
    cin >> n >> k;
    vector<long long> aa(n); 

    for (int i = 0; i < n; i++) {
        cin >> aa[i];
    }

    sort(aa.begin(), aa.end()); 

    if (k <= n) {
        cout << aa[k - 1] << endl;
    }

    return 0;
}

2.归并排序

……

3.二分

……

4.高精度

4.1高精度加法

#include <iostream>
#include <vector>

using namespace std;
const int N = 10e6+10;
vector<int> add(vector<int> &A, vector<int> &B)  
{
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i ++ )
    {
        
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % 10);
        t /= 10;
    }

    if (t) C.push_back(1);
    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');
    
    auto C = add(A,B);
    for(int i = C.size()-1;i>=0;i--) {
        printf("%d",C[i]);
    }
    return 0;    
}

4.2高精度减法

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

using namespace std;
bool cmp (vector<int> &A, vector<int> &B){
    if (A.size()!=B.size()) return A.size() > B.size();
    else{
        for (int i = A.size() - 1;i>=0;i--){
            if(A[i]!=B[i]){
             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;
        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;
}

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');
    if(cmp(A,B)){
        auto C = sub(A,B);
    for(int i = C.size()-1;i>=0;i--) cout<< C[i];
    }
    else{
          auto C = sub(B,A);
          cout<<"-";
    for(int i = C.size()-1;i>=0;i--)cout<< C[i];
    }
    return 0;
}

4.3高精度乘法

4.3.1大数 * 小数

#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();

    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--){
        cout<< C[i];
    }  
    return 0;
}

4.3.2大数 * 大数

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> A, vector<int> B)
{
    vector<int> C(A.size() + B.size());

    for (int i fo'r= 0; i < A.size(); i ++ )
        for (int j = 0; j < B.size(); j ++ )
            C[i + j] += A[i] * B[j];

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

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

    return C;
}

int main()
{
    string a, b;
    cin >> a >> b;
    vector<int> 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');
    auto C = mul(A, B);

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

    return 0;
}

4.4高精度除法

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

using namespace std;
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;
}

int main()
{
    string a;
    int b;
    int r;
    cin>>a>>b;
    vector<int> A;
    for(int i = a.size()-1;i>=0;i--){
        A.push_back(a[i]-'0');
    }
    auto C = div(A,b,r);
    
    for(int i = C.size()-1;i>=0;i--){
        cout << C[i];
    }
    cout<<endl<<r;
}

5.前缀和与差分

5.1前缀和

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 100010;
int m,n;
int a[N],s[N];
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 ++ ){
        s[i] = s[i-1] + a[i];
    }
    while (m -- ){
        int l,r;
        scanf("%d%d", &l, &r);
        cout<<s[r]-s[l-1]<<endl;
    }
}

5.2子矩阵的和(二维前缀和)

5.2.1如何计算S[i,j]

S[i][j] = S[i - 1][j] + S[i][j- 1] - S[i - 1][j- 1] + a[i][j]

5.2.2如何计算(x1,y1),(x2,y2)这一子矩阵中所有数的和

S[x2][y2] - S[x1 - 1][y2] - S[x2][y1 - 1] + S[x1 - 1][y1 - 1]

#include <iostream>

using namespace std;
const int N = 1010;
int n,m,q;
int S[N][N],a[N][N];
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 ++ ){
           S[i][j] = S[i - 1][j] + S[i][j- 1] - S[i - 1][j- 1] + a[i][j];
        }
    }      
    while(q--){
        int x1,y1,x2,y2;
        scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
        printf("%d\n",S[x2][y2] - S[x1 - 1][y2] - S[x2][y1 - 1] + S[x1 - 1][y1 - 1]);
    }
}

5.3.差分

5.4差分矩阵