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…

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…

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…

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…

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 …

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…

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>…

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.…