快速幂递归快速幂//递归快速幂 ll qpow(int a,int n) {if(n == 0) {return 1;}else if (n%2 == 1) {return qpow(a,n-1) * a;}else {long long temp = qpow(a,n/2);return temp

取余运算的性质同余式性质逆元:a⋅b≡1(mod p)称a是b的乘法逆元，逆元: a·b ≡ 1 (mod~p) \\称a是b的乘法逆元，逆元:a⋅b≡1(mod p)称a是b的乘法逆元，那如何计算逆元呢？使用快速幂算法就可以了int qpow(int a,int n,int p) { int

#include <iostream>#include <vector>#include <stdio.h>using namespace std;class TreeNode {public: int val; TreeNode *left;

模板/* 滑动窗口算法框架 */void slidingWindow(string s, string t) { unordered_map<char, int> window; int left = 0, right = 0; while (right < s.

https://blog.csdn.net/qq_36102055/article/details/104375766https://www.acwing.com/solution/content/107848/https://www.acwing.com/problem/content/791/

https://www.acwing.com/solution/content/218914/https://www.acwing.com/solution/content/218913/

#include <iostream>#include <cstring>int dp(int memo[],int n){if(n == 0 || n == 1) return n; if(memo[n] != 0) return memo[n]; memo[n

#ifndef BINARYTREE_H#define BINARYTREE_H#pragma once#include <iostream>#include <cassert>#include <vector>#include <stack>#inc

https://www.acwing.com/problem/content/797/#include <iostream>using namespace std;const int N = 100010;int n, m;int a[N], s[N];int main(void){

分解质因数1、题目：给定n个正整数ai，将每个数分解质因数，并按照质因数从小到大的顺序输出每个质因数的底数和指数。输入格式第一行包含整数n。接下来n行，每行包含一个正整数ai。输出格式对于每个正整数ai,按照从小到大的顺序输出其分解质因数后，每个质因数的底数和指数，每个底数和指数占一行。每个正整数的