The Last Practice
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8374 Accepted Submission(s): 1724
Problem Description
Tomorrow is contest day, Are you all ready?
We have been training for 45 days, and all guys must be tired.But , you are so lucky comparing with many excellent boys who have no chance to attend the Province-Final.
Now, your task is relaxing yourself and making the last practice. I guess that at least there are 2 problems which are easier than this problem.
what does this problem describe?
Give you a positive integer, please split it to some prime numbers, and you can got it through sample input and sample output.
Input
Input file contains multiple test case, each case consists of a positive integer n(1<n<65536), one per line. a negative terminates the input, and it should not to be processed.
Output
For each test case you should output its factor as sample output (prime factor must come forth ascending ), there is a blank line between outputs.
Sample Input
60 12 -1
Sample Output
Case 1. 2 2 3 1 5 1 Case 2. 2 2 3 1 Hint 60=2^2*3^1*5^1
Author
lcy
这个题目气的我头疼，以前做过类似的题目 ，思路先是 素数打表，然后依次检索并记录素数因子，重复此过程直到，n==1；不过不幸的是超时。参考了网上的方法。
不好理解，可以这样想。一个数既然可以表示某些素数的乘积，那么对于一个数n，从2开始检索n的所有因子，那么既然是n的因子，必然为质数，反证：如果某个i为合数，
那么必然又能写成两小一点儿得数乘积，早就应该被从2开始的因子分解掉了，不可能发生，那么只能是质数了。
这个是超时的：
#include<iostream>
#include<cmath>
#include<string.h>
using namespace std;
int ls[65539];
int cnt[1000];
int u=0;
void prime()
{
for(int i=1;i<=65537;i++)
ls[i]=1;
for(int j=2;j<sqrt(65537);j++)
{
if(ls[j])
{
for(int k=j+j;k<=65537;k+=j)
ls[k]=0;
}
}
for(int r=2;r<=65537;r++)
{
if(ls[r])
cnt[u++]=r;
}
}
int main()
{
int n;int gq[65540];int h=1;
prime();
while(cin>>n&&n>0)
{
memset(gq,0,sizeof(gq));
for(int x=0;x<u;x++)
{
if(n%cnt[x]==0)
{
gq[cnt[x]]++;
n/=cnt[x];
}
if(n==1)
break;
else if(n<=cnt[x])
x=-1;
}
cout<<"Case "<<h++<<"."<<endl;
for(int p=0;p<=65534;p++)
{
if(gq[p])
cout<<p<<" "<<gq[p]<<" ";
}
cout<<endl<<endl;
}
return 0;
}
AC Code：
#include<iostream>
using namespace std;
int main()
{
int n,ls[100],gq[50],flag=0,t=1;
while(cin>>n&&n>0)
{
if(flag)
cout<<endl;
memset(ls,0,sizeof(ls));
memset(gq,0,sizeof(gq));
int k=0;
for(int i=2;n!=1;i++)
{
if(n%i==0)
{
ls[k]=i;
while(n%i==0)
{
gq[k]++;
n/=i;
}
k++;
}
}
cout<<"Case "<<t++<<"."<<endl;
for(int m=0;m<k;m++)
cout<<ls[m]<<" "<<gq[m]<<" ";
cout<<endl;
flag=1;
}
return 0;
}