基础算法进阶_基本运算的进阶-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_45382931/article/details/122320368

lowbit

非负整数n在二进制表示的形式，最低为的1和以及最低位1后的所有0组成的数
lowbit(n) = n & (-n)

快速幂

typedef long long LL;

int qmi(int a, int k, int p)  // 求a^k mod p
{
    //这一步不能少
    a %= p;
    int res = 1 % p;
    while (k)
    {
        if (k & 1) res = (LL)res * a % p;
        a = (LL)a * a % p;
        k >>= 1;
    }
    return res;
}

约数之和(acwing 97)

题意：求 $A^B$ 的所有约数之和
分析： $A可以分解质因数变成a_{1}^{b1} * a_{2}^{b2} * ...* a_{n}^{bn}$ ,
$则A^B分解质因数可以变成a_{1}^{B * b1}* a_{2}^{B * b2}*...*a_{n}^{B * bn}$ ； $因为a_{1},...,a_{n}是A^B的约数，所以a_{1}^{b1}*...*a_{n}^{bn}也是A^B的约数$ ；
$a_1^{b1}的取法有b1 + 1种，例如a_1^0,a_1^1,...,a_1^{b1}$ ;
$因此总的约数个数之和为 (b 1 + 1) * (b 2 + 1) * . . . * (b n + 1)$ ;
$总的约数之和为(a_{1}^0 + a_{1}^1 +...+a_1^{b1})*...*(a_{n}^0 + a_{n}^1 +..+a_n^{bn})$ ;
$以sum(a, k)代表(a^0 + a ^ 1 + ... + a^k)$
$若 k 为奇数, 则 k + 1 为偶数，有 k + 1 个约数 (0 - k)$
$^0 + ... + a^\frac{k-1}{2}) + (a^\frac{k+1}{2} + ...+ a^k)$
$^0 + ... + a^\frac{k-1}{2}) + a^\frac{k+1}{2} (a^0+ ...+ a^\frac{k-1}{2})$
$(1+a^\frac{k+1}{2})*sum(a, \frac{k-1}{2})$
$若 k 为偶数, 则 k + 1 为奇数, k - 1 为偶数$
$则sum(a, k) = sum(a, k - 1) + a^k$
代码：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int mod = 9901;
//快速幂
int qmi(int a, int k, int p)  // 求a^k mod p
{
    //这一步不能少
    a %= p;
    int res = 1 % p;
    while (k)
    {
        if (k & 1) res = res * a % p;
        a = a * a % p;
        k >>= 1;
    }
    return res;
}


int sum(int a, int k){
    //结束条件,a^0=1
    if(k == 0) return 1;
    if(k % 2 == 1){
        //k取余2等于1，说明k是奇数，即有偶数个约数
        //sum(a,k) = (1+a^(k+1/2))*sum(a, (k-1/2))
        return (1 + qmi(a, (k + 1) / 2, mod)) * sum(a, (k - 1) / 2) % mod;  
    }
    //sum(a, k) = sum(a, k - 1) + a^k
    return (sum(a, k - 1) + qmi(a, k, mod)) % mod;
}

int main()
{
	int A, B;
    cin >> A >> B;
    //分解质因数
    int res = 1;
    for(int a = 2; a <= A; a ++){
        int k = 0;
        while(A % a == 0){
            A /= a;
            k ++;
        }
        if(k > 0){
            //求解sum(a,k)
            res = res * sum(a, B * k) % mod;
        }
    }
    if(A == 0) res = 0;
    cout << res;
}

前缀和

$\sum_{1<=j<=i}^{i}a[j]$

差分

$d [1] = a [1]$
$d [i] = a [i] - a [i - 1] (i > = 2)$
差分还原
$a [i] = a [i - 1] + d [i]$

二分

前驱二分

while(l < r){
	int mid = (l + r + 1) >> 1;
	if(a[mid] <= x) l = mid; else r = mid - 1;  
}
return a[l];

后继二分

while(l < r){
	int mid = l + r >> 1;
	if(a[mid] >= x) r = mid; else l = mid + 1;
}
return a[l];

实数域二分

//保留k位小数
int k = 3;
double esp = 10^-(k + 2);
while(l + esp < r){
	double mid = (l + r) / 2;
	if(calc(mid)) r = mid; else l = mid;
}

深度优先遍历

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

using namespace std;

const int N = 2 * 1e5 + 10;
int e[N], ne[N], h[N], idx;
int st[N];
int n;

void add(int a, int b){
    e[idx] = b,ne[idx] = h[a], h[a] = idx++; 
}

void dfs(int x){
    //前序遍历
    cout << x;
    st[x] = 1;
    for (int i = h[x]; i != -1; i = ne[i] ){
        int val = e[i];
        if(st[val] == 1) continue;
        dfs(val);
    }
    //后序遍历
    cout << x;
}

int main()
{
    //初始化
    memset(h, -1, sizeof h);
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int a,b;
        scanf("%d%d", &a, &b);
        add(a, b);
        add(b, a);
    }
    dfs(1);
}

树的深度

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

using namespace std;

const int N = 2 * 1e5 + 10;
int e[N], ne[N], h[N], idx;
int st[N], deep[N];
int n;

void add(int a, int b){
    e[idx] = b,ne[idx] = h[a], h[a] = idx++; 
}

void dfs(int x){
    st[x] = 1;
    for (int i = h[x]; i != -1; i = ne[i] ){
        int val = e[i];
        if(st[val] == 1) continue;
        //子节点深度是父节点的深度+1
        deep[val] = deep[x] + 1;
        dfs(val);
    }
}

int main()
{
    //初始化
    memset(h, -1, sizeof h);
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int a,b;
        scanf("%d%d", &a, &b);
        add(a, b);
        add(b, a);
    }
    dfs(1);
    cout<< deep[n];
}

树的重心

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

using namespace std;

const int N = 2 * 1e5 + 10;
int e[N], ne[N], h[N], idx;
int st[N], siz[N];
int n, ans = 1e9, pos;

void add(int a, int b){
    e[idx] = b,ne[idx] = h[a], h[a] = idx++; 
}

void dfs(int x){
    st[x] = 1;
    siz[x] = 1;
    int maxNode = 0; //最大子树大小 
    for (int i = h[x]; i != -1; i = ne[i] ){
        int val = e[i];
        if(st[val] == 1) continue;
        dfs(val);
        siz[x] += siz[val];
        maxNode = max(maxNode, siz[val]);
    }
    maxNode = max(maxNode, n - siz[x]);
    if(maxNode < ans){
        ans = maxNode; //最大子节点
        pos = x; //重心
    }
}

int main()
{
    //初始化
    memset(h, -1, sizeof h);
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ){
        int a,b;
        scanf("%d%d", &a, &b);
        add(a, b);
        add(b, a);
    }
    dfs(1);
    cout<< pos;
}

广度优先遍历

在这里插入图片描述

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

using namespace std;
const int N = 2*1e5 + 10;

int e[N], ne[N], h[N], idx;
queue<int> q;
bool st[N];
void add(int a, int b){
    e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void bfs(){
    memset(st, false, sizeof st);
    queue<int> q;
    q.push(1);
    st[1] = true;
    while (q.size() > 0){
        int x = q.front();
        cout << x << endl;
        q.pop();
        for (int i = h[x]; i != -1; i = ne[i]){
            int j = e[i];
            if(st[j]) continue;
            st[j] = true;
            q.push(j);
        }
    }
}


int main()
{
    memset(h, -1, sizeof h);
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++ ){
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    bfs();
}

拓扑排序

在这里插入图片描述

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

using namespace std;

const int N = 1e5 + 10;
int e[N], ne[N], h[N], idx;
int d[N], st[N];

int n,m;

void add(int a, int b){
    //a指向b, b入度+1
    d[b] += 1;
    e[idx] = b, ne[idx]= h[a], h[a] = idx++;
}

void bfs(){
    vector<int> v;
    queue<int> q;
    //入度为0的点入队
    for (int i = 1; i <= n; i ++ ){
        if(d[i] == 0) q.push(i);
    }
    while (q.size() > 0){
        int x = q.front();
        //被访问过的元素打个标记
        st[x] = true; 
        v.push_back(x);
        q.pop();
        for (int i = h[x]; i != -1; i = ne[i] ){
            int y = e[i];
            if(st[y]) continue;
            //该点指向的所有点入度减1
            d[y] -= 1;
            //若入度为0，入队
            if(d[y] == 0) {
                q.push(y);
            }
        } 
    }
    if(v.size() != n) printf("%d", -1);
    else{
        for (int i = 0; i < v.size(); i ++ )printf("%d ", v[i]);
    }
}

int main()
{
    memset(h, -1, sizeof h);
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i ++ ){
        int a, b;
        scanf("%d%d", &a, &b);
        add(a, b);
    }
    bfs();
}