算法基础课第一章基础算法

Posted 2021-12-28 红叶~

快排

void quick_sort(int q[], int l, int r)

    if (l >= r) return;

    int i = l - 1, j = r + 1, x = q[l + r >> 1];
    while (i < j)
    
        while (q[++ i] < x);
        while (q[-- j] > x);
        if (i < j) swap(q[i], q[j]);
    
    quick_sort(q, l, j), quick_sort(q, j + 1, r);

二分

整数二分

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;

浮点数二分

// 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;

位运算

求n的第k位数字: n >> k & 1

返回n的最后一位1：lowbit(n) = n & -n	x -= lowbit(n)  // 去除n的最后一位1

归并排序

#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;

高精度

加法

t+=a[i]+b[i]; c.push_back(t%10); t/=10

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

vector<int> add(vector<int> &a,vector<int> &b)
    vector<int> c;
    int t=0;//进位
    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;//进位权重下降
    
    if(t) c.push_back(1);
    return c;


int main()
    vector<int> a,b;
    string 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=add(a,b);
    for(int i=c.size()-1;i>=0;i--) printf("%d",c[i]);
    return 0;

减法

利用cmp比较大小

t=a[i]-t;t -=b[i];if (t<0); t=1;else t = 0

#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();

    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\');

    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;

乘法

t+=a[i]*b;c.push_back(t%10);t/=10

高精乘低精

#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 -- ) printf("%d", C[i]);

    return 0;

高精乘高精

#include <iostream>
#include <vector>

using namespace std;

vector<int> mul(vector<int> &A, vector<int> &B) 
    vector<int> C(A.size() + B.size(), 0); // 初始化为 0，且999*99最多 5 位

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

    int t = 0;
    for (int i = 0; i < C.size(); i++)  // i = C.size() - 1时 t 一定小于 10
        t += C[i];
        C[i] = t % 10;
        t /= 10; 
    

    while (C.size() > 1 && C.back() == 0) C.pop_back(); // 必须要去前导 0，因为最高位很可能是 0
    return C;


int main() 
    string a, b;
    cin >> a >> b; // a = "1222323", b = "2323423423"

    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];

    return 0;

除法

r=r*10+ a[i];c.push_back(r/b);r%=b

#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 -- ) printf("%d", C[i]);

    return 0;

前缀和差分

..................

离散化

vectotr<int> alls;存储所有待离散化的值

sort(alls.begin(),alls.end());将所有值排序

alls.erase(unique(alls.begin(), alls.end()), alls.end())去除重复元素

// 二分求出x对应的离散化的值
int find(int x) 
    int l = 0, r = alls.size() - 1;
    while(l < r) 
        int mid = l + r >> 1;
        if (alls[mid] >= x)	r = mid;
        else l = mid + 1;
    
    return r + l; // 如果要求前缀和下标映射到1,2,3...n

快速幂

int main() 
    long long a, b, ans = 1, k;
    cin >> a >> b >> k;
    while(b) 
        if (b & 1)	ans = ans * a % k;
        a = a * a % k;
        b >>= 1;
    
    if(k == 1)	ans = 0;
    cout << a