1059. Prime Factors (25)

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;//因为题目说在int范围内（10^9）的正整数进行质因子分解，因此素数表大概开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;        }    }}//本来想用个HashTable[],下标存放质因子，值存其个数，如HashTable[5]=4,即4个5，//但输出时不方便，必须开一个int范围的数组（n=10^9)，且要遍历全部元素， struct factor{     int x,cnt;//x为质因子，x为其个数 }fac[10];//int型（10^9)范围内的数，不同质因子数不可能超过10个，        //因为对于x=2*3*5*7*11*13*17*19*23*29,x有十个质因子，x已超出int范围，即一个数若有十个不同质因子，它一定大于等于x,超出int范围 int main(){    findPrime();    int n,num=0;//num为n的不同质因子个数     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){//如果无法被根号n以内的质因数除尽，如38=2*19，根号n为6<x<7,循环结束时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;}