标签：
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 long int.
Output Specification:
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 one pi, ki is 1 and must NOT be printed out.
Sample Input:
97532468
Sample Output:
97532468=2^2*11*17*101*1291
思路:此题真是受益匪浅，首先需要观察素数表的打印，然后通过建立结构体进而来存储因子，而结构体中还可以存储数字。另外特殊的情况需要进行考虑。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 using namespace std; 5 #define MAX 100010 6 struct Fraction 7 { 8 int num; 9 int cnt; 10 } Frac[10]; 11 bool Judge(int a) 12 { 13 if(a<=1) 14 return false; 15 int sqr= sqrt(a*1.0); 16 for(int i=2;i<=sqr;i++) 17 { 18 if(a%i==0) 19 return false; 20 } 21 return true; 22 } 23 int prime[MAX]; 24 void Init() 25 { 26 27 //需要打印素数表 28 int npt=0; 29 for(int i=2;i<MAX;i++) 30 { 31 if(Judge(i)) 32 prime[npt++]=i; 33 } 34 } 35 int main(int argc, char *argv[]) 36 { 37 int n; 38 scanf("%d",&n); 39 printf("%d=",n); 40 Init(); 41 int i=0; 42 int pt=0; 43 if(n==1) 44 { 45 printf("1\n"); 46 return 0; 47 } 48 int sqr=sqrt(1.0*n); 49 for(int i=0;i<MAX&&prime[i]<=sqr;i++) 50 { 51 if(n%prime[i]==0) 52 { 53 Frac[pt].num=prime[i]; 54 Frac[pt].cnt=0; 55 while(n%prime[i]==0) 56 { 57 Frac[pt].cnt++; 58 n/=prime[i]; 59 } 60 pt++; 61 } 62 } 63 for(int i=0;i<pt;i++) 64 { 65 if(Frac[i].cnt!=0) 66 { 67 printf("%d",Frac[i].num); 68 if(Frac[i].cnt!=1) 69 { 70 printf("^%d",Frac[i].cnt); 71 } 72 if(i!=pt-1) 73 putchar(‘*‘); 74 else if(n==1) 75 putchar(‘\n‘); 76 } 77 } 78 if(n>1) 79 { 80 printf("%d\n",n); 81 } 82 return 0; 83 }
标签：
原文地址：http://www.cnblogs.com/GoFly/p/4296902.html
