TOJ 10004 Prime Factorization

题目链接 : TOJ10004

10004 - Prime Factorization

Time Limit: 1000MS
Memory Limit: 65536KB

 

Description
For a given integer, "Prime Factorization" is finding which prime numbers you need to multiply together to get the original number, also called prime decomposition.

Given a positive integer n>=2, the prime factorization is written

n = p₁^a₁p₂^a₂...p_k^a_k

where the p_is are thek prime factors, each of order a_i.

Write a program to factor a given integer n.



Input
One integer, n. (2<=n<=2³¹-1).



Output
Output the prime factorization of n as follows:

n=p₁(a₁)p₂(a₂)...p_k(a_k)

Where p₁<p₂<...<p_k, and a_i denotes the exponent of p_i.

For example, for n = 1500 = 2²*3*5³, the output should be:

1500=2(2)3(1)5(3)



Sample Input #1
2


Sample Output #1
2=2(1)



Sample Input #2
15


Sample Output #2
15=3(1)5(1)



Sample Input #3
1001


Sample Output #3
1001=7(1)11(1)13(1)



Note
Take a look at problem 20002 if you're interested in integer factoring problems.

大数的素因子分解 

先筛小素数  然后用试除法 

在O(sqrtn) 的判断试除后的数是不是质数

 

#include<cstdio>bool vis[1000005];int prim[90000];int con[90000];int cal(){int k=0;for(int i=2;i<1000001;i++){if(vis[i]==0) prim[k++]=i;for(int j=0;prim[j]*i<1000001 && j<k;j++){vis[i*prim[j]]=1;if(i%prim[j]==0) break;}}return k;}bool is(int n){bool l=1;for(long long i=2;i*i<=n;i++){if(n%i==0) {l=0;break;}}return l;}int main(){int n,num=cal(),a=0,sp=0,ans,max=0;scanf("%d",&n);printf("%d=",n);while(n!=1){if(is(n)) {sp=n;break;}int c=n;while(c==n){while(n%prim[a]==0){max=a;n/=prim[a];con[a]++;}a++;}}for(int i=0;i<=a;i++){if(con[i]){printf("%d(%d)",prim[i],con[i]);}}if(sp) printf("%d(1)\n",sp);else putchar(10);return 0;}