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
#include<cstdio>#include<cmath>const int maxn=100010;//因?yàn)轭}目說在int范圍內(nèi)（10^9）的正整數(shù)進(jìn)行質(zhì)因子分解，因此素?cái)?shù)表大概開10^5就可以了 bool isPrime(int x){ bool flag=true; if(x<=1) flag=false; for(int i=2;i*i<=x;i++){ if(x%i==0){ flag=false; break; } } return flag;}int prime[maxn],pNum=0;void findPrime(){ for(int i=2;i<maxn;i++){ if(isPrime(i)){ prime[pNum++]=i; } }}//本來想用個(gè)HashTable[],下標(biāo)存放質(zhì)因子，值存其個(gè)數(shù)，如HashTable[5]=4,即4個(gè)5，//但輸出時(shí)不方便，必須開一個(gè)int范圍的數(shù)組（n=10^9)，且要遍歷全部元素， struct factor{ int x,cnt;//x為質(zhì)因子，x為其個(gè)數(shù) }fac[10];//int型（10^9)范圍內(nèi)的數(shù)，不同質(zhì)因子數(shù)不可能超過10個(gè)， //因?yàn)閷?duì)于x=2*3*5*7*11*13*17*19*23*29,x有十個(gè)質(zhì)因子，x已超出int范圍，即一個(gè)數(shù)若有十個(gè)不同質(zhì)因子，它一定大于等于x,超出int范圍 int main(){ findPrime(); int n,num=0;//num為n的不同質(zhì)因子個(gè)數(shù) scanf("%d",&n); if(n==1) printf("1=1/n"); else{ printf("%d=",n); int sqr=(int)sqrt(n); 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){//如果無法被根號(hào)n以內(nèi)的質(zhì)因數(shù)除盡，如38=2*19，根號(hào)n為6<x<7,循環(huán)結(jié)束時(shí)n=19 fac[num].x=n; fac[num++].cnt=1; } for(int i=0;i<num;i++){ printf("%d",fac[i].x); if(fac[i].cnt>1){ printf("^%d",fac[i].cnt); } if(i<num-1) printf("*"); } } return 0;}新聞熱點(diǎn)
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