1059 Prime Factors (25 分) Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specification: Each input file contains one test case which…

题目链接 http://www.patest.cn/contests/pat-a-practise/1059 Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km. Input Specification: Each input file contains one test ca…

反正知道了就是知道,不知道也想不到,很快 #include <cstdio> #include <cstdlib> #include <vector> using namespace std; inline void print_prime_k(long long p, long long k) { printf("%lld", p); ) { printf("^%lld", k); } } int main() { long l…

1059 Prime Factors (25 分)   Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specification: Each input file contains one test case whi…

1059 Prime Factors (25分) 1. 题目 2. 思路 先求解出int范围内的所有素数,把输入x分别对素数表中素数取余,判断是否为0,如果为0继续除该素数知道余数不是0,遍历到sqrt(x)就够了 3. 注意点 输入数为1的情况, 输出1=1 4. 代码 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; #d…

题目 Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 -pm^km. Input Specification: Each input file contains one test case which gives a positive integer N in the range of lo…

https://pintia.cn/problem-sets/994805342720868352/problems/994805415005503488 Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specifi…

时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1* p2^k2 *…*pm^km. Input Specification: Each input file contain…

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specification: Each input file contains one test case which gives a positive integer …

用素数筛选法即可. 范围long int,其实大小范围和int一样,一开始以为是指long long,想这就麻烦了该怎么弄. 而现在其实就是int的范围,那难度档次就不一样了,瞬间变成水题一枚,因为int根号后大小不超过60000. 即因子的大小不会超过60000,除非本身是质数. int 4 -2147438648-+2147438647long int 4 -2147438648-+2141438647 #include <iostream> #include <cstdio>…

素因子分解. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<stack> #include<vector> using namespace st…

题意: 输入一个正整数N(范围为long int),输出它等于哪些质数的乘积. trick: 如果N为1,直接输出1即可,数据点3存在这样的数据. 如果N本身是一个质数,直接输出它等于自己即可,数据点4存在这样的数据. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int main(){ ios::sync_with_stdio(false); ci…

题意: 给出一个int型正整数N,要求把N分解成若干个质因子,如N=97532468,则把N分解成:97532468=2^2*11*17*101*1291.质因子按增序输出,如果某个质因子的幂是1,则1不输出. 思路:质因子分解的基础题. 首先,定义如下因子的结构体,用于存放最终的结果.因为N是一个int范围的正整数,由于2*3*5*7*11*13*17*19*23*29>INT_MAX,也就是说,任意一个int型整数,可分解出来的不同的质因子的个数不会超过10个,因此,数组只要开到10就够了.…

1059. Prime Factors Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *-*pm^km. Input Specification: Each input file contains one test case which gives a positive integer N…

1059. Prime Factors (25) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *-*pm^km. Input Specificati…

Source: PAT A1059 Prime Factors (25 分) Description: Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specification: Each input file co…

1059. Prime Factors (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km. Input Specificatio…

1059. Prime Factors (25) 时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 HE, Qinming Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km. Input Specificatio…

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p​1​​​k​1​​​​×p​2​​​k​2​​​​×⋯×p​m​​​k​m​​​​. Input Specification: Each input file contains one test case which gives a positive integer …

Largest prime factor Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13481    Accepted Submission(s): 4765 Problem Description Everybody knows any number can be combined by the prime number.Now,…

7.7 Design an algorithm to find the kth number such that the only prime factors are 3,5, and 7. 这道题跟之前LeetCode的那道Ugly Number II 丑陋数之二基本没有啥区别,具体讲解可参见那篇,代码如下: class Solution { public: int getKthMagicNumber(int k) { vector<, ); , i5 = , i7 = ; while (re…

Given any positive integer N, you are supposed to find all of its prime factors, and write them in the format N = p1^k1 * p2^k2 *…*pm^km. Input Specification: Each input file contains one test case which gives a positive integer N in the range of lon…

Problem Description Everybody knows any number can be combined by the prime number. Now, your task is telling me what position of the largest prime factor. The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc. Specially, LPF(1) = 0. Inpu…

问题 L: Prime Factors 时间限制: 1 Sec  内存限制: 128 MB [提交][状态][论坛] 题目描写叙述 I'll give you a number , please tell me how many different prime factors in this number. 输入 There is multiple test cases , in each test case there is only one line contains a number N(…

本题涉及博弈论中的Nim游戏博弈. Nim游戏博弈详解链接: http://www.cnblogs.com/exponent/articles/2141477.html 本题解题报告详解链接: http://blog.csdn.net/woshi250hua/article/details/7824609 我的代码,写的较挫,4000多ms,压着时间过,时间限制是5s.我预处理了所有1-5*10^6的数的素因子个数,用的是dp 的思想,先预处理1-5*10^6的数的最小素因子,存在a数组中,然后…

Prime Factorization using Sieve O(log n) for multiple queries 使用筛法在 O(logN) 的时间内查询多组数的素数因子 前言 通常, 我们使用 O(n ^ 2) 的两层循环来进行打表, 记录一个数字是否为素数, 再使用 O(n) 的循环来求所有素因子. 然而, Nitish Kumar提供了一种在 O(nloglogn) 内打表, O(logN) 时间内查询的算法. 核心思想 这种算法的核心思想是, 使用 spf[] 存放每一个数的最…

PAT 1059. C语言竞赛 C语言竞赛是浙江大学计算机学院主持的一个欢乐的竞赛.既然竞赛主旨是为了好玩,颁奖规则也就制定得很滑稽: 冠军将赢得一份"神秘大奖"(比如很巨大的一本学生研究论文集--). 排名为素数的学生将赢得最好的奖品 -- 小黄人玩偶! 其他人将得到巧克力. 给定比赛的最终排名以及一系列参赛者的ID,你要给出这些参赛者应该获得的奖品. 输入格式: 输入第一行给出一个正整数N(<=10000),是参赛者人数.随后N行给出最终排名,每行按排名顺序给出一位参赛者的I…

埃拉托斯特尼筛法(sieve of Eratosthenes ) 是古希腊数学家埃拉托斯特尼发明的计算素数的方法.对于求解不大于n的所有素数,我们先找出sqrt(n)内的所有素数p1到pk,其中k = sqrt(n),依次剔除Pi的倍数,剩下的所有数都是素数. 具体操作如上述 图片所示. C++实现 #include<iostream> #include<vector> using namespace std; int main() { int n; cin >> n;…

题意: 给出一个n ; 有两个操作: 1,mul A   ,   n=n*A   : 2,sqrt()  ,  n=sqrt(n)  开更出来必须是整数 : 求出经过这些操作后得出的最小  n , 和最小操作数: 分析:首先得明确知道分解到怎样的时候才是 得出最小的n , 首先进过手画就可以明明,经过分解n 可以发现它的素数因子是不可以被开根去的 , 无论怎么相乘开根的结果总是留在的 , 好那这到题的MIN(n) = (本身所有的素数因子的鸡) : 那操作数怎么求呢? 首先我们只需要一次的相乘把…

POJ 3518 Prime Gap(素数) id=3518">http://poj.org/problem? id=3518 题意: 给你一个数.假设该数是素数就输出0. 否则输出比这个数大的素数与比这个数小的素数的差值. 分析: 明显本题先要用筛选法求出130W(个素数)以内的全部素数. 然后推断给的数是否是素数就可以. 假设不是素数.那么就找出它在素数素组内的上界和下界,输出两个素数的差值就可以. 筛选法求素数可见: http://blog.csdn.net/u013480600/a…