PAT(甲级)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 gives a positive integer N in the range of long int.

Output Specification:

Factor N in the format N = p​1​​^k​1​​*p​2​​^k​2​​*…*p​m​​^k​m​​, where p​i​​'s are prime factors of N in increasing order, and the exponent k​i​​ is the number of p​i​​ -- hence when there is only one p​i​​, k​i​​ is 1 and must NOT be printed out.

Sample Input:

97532468

Sample Output:

97532468=2^2*11*17*101*1291

简析：就是把一个int型的数用它的最小约数来表示，这道题主要是结合素数的寻找。简单的方法就是先把小于这个数开方所得的数的所有素数给先统计出来。再根据最小的10个素数相乘所得的数字已经超过了int型数的最大值，那么最多的输出素数就是10个，可以设置10个结构体数组，来存放组成所给数的约数和它的数量。 最后从小到大挨个输出就可以了。

#include<iostream> #include<cmath> using namespace std; const int maxn = 100010; bool is_prime(int n) { if(n == 1) return false; int sqr = (int)sqrt(n * 1.0); for (int i = 2; i <= sqr;i++) { if(n%i == 0) return false; } return true; } int prime[maxn], pNum = 0; void Find_prime() { for (int i = 1; i < maxn;i++) { if(is_prime(i) == true) prime[pNum++] = i; } } struct factor{ int x, cnt; } fac[10]; int main() { Find_prime(); int n,num = 0; cin >> n; if(n == 1) printf("1=1"); else{ printf("%d=", n); int sqr = (int)sqrt(n * 1.0); for (int i = 0; i < pNum && prime[i] <= sqr;i++) { if(n % prime[i] == 0) { fac[num].x = prime[i]; fac[num].cnt = 0; while(n % prime[i] == 0) { fac[num].cnt++; n /= prime[i]; } num++; } if(n == 1) break; } if(n != 1){ fac[num].x = n; fac[num++].cnt = 1; } for (int i = 0; i < num;i++) { if(i>0) cout << '*'; cout << fac[i].x; if(fac[i].cnt > 1) cout << '^' << fac[i].cnt; } } return 0; }