Factor N in the format N = p1^k1* p2^k2*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent kiis the number of pi-- hence when there is only one pi, kiis 1 and must NOT be printed out. Sample Input: 97532468 Sample Output: 97532468=2^2*11*1...
Factor N in the format N = p1^k1* p2^k2*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent kiis the number of pi-- hence when there is only one pi, kiis 1 and must NOT be printed out. Sample Input: 97532468 Sample Output: 97532468=2^2*11*1...
简介: 【PAT甲级 - C++题解】1059 Prime Factors 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 ...
PAT 甲级 1059 Prime Factors (25 分) 题目描述 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 3 k 3 ∗ . . . ∗ p m k m p_1^{k1......
当n%prime[i] == 0时 说明是质因子,(一个数有多个不同的质因子) 设定factor[i]结构体存放 不同的质因子 并记录他们的个数 while(n % prime[i] == 0)时 质因子个数 + 1, n = n / prime[i]; **/ 具体代码: #include<iostream>
PAT:1059. Prime Factors (25) AC #include<stdio.h> #include<math.h> #include<string.h> #include<algorithm> using namespace std; const int MAX=100010; //int型素数一定在这个范围内 int PrimeArr[MAX]; //素数表 int cnt=0; //素数表中素数个数 int x,x2; //输入的数字,x2是x的开平方...
#include <cstdio> #include <vector> using namespace std; vector<int> prime(50000, 1); int main() { for(int i = 2; i * i < 50000; i++) for(int j = 2; j * i < 50000; j++) prime[j * i] = 0; long int a; scanf("%ld", &a); printf("%ld=", a); if(a == ...
FactorNin the formatN=p1^k1*p2^k2*…*pm^km, wherepi's are prime factors ofNin increasing order, and the exponentki is the number ofpi -- hence when there is only onepi,ki is 1 and mustNOTbe printed out. ...
if(n%prime[i]==0) {//如果prime[i]是n的因子 fac[num].x=prime[i];//记录该因子 fac[num].cnt=0; while(n%prime[i]==0) {//计算出质因子prime[i]的个数 fac[num].cnt++; n/=prime[i]; } num++;//不同质因子个数加1
Factor N in the format N=p1^k1*p2^k2*…*pm^km, where pi's are prime factors of N in increasing order, and the exponent ki is the number of pi -- hence when there is only on...