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…

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…

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…

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…

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…

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…

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…

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…