Largest prime factor
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Largest prime factor
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5632 Accepted Submission(s): 1712Problem Description
Everybody knows any number can be combined by the prime number.
Now, your task is telling me what position of the largest prime factor.
The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.
Specially, LPF(1) = 0.
Now, your task is telling me what position of the largest prime factor.
The position of prime 2 is 1, prime 3 is 2, and prime 5 is 3, etc.
Specially, LPF(1) = 0.
Input
Each line will contain one integer n(0 < n < 1000000).
Output
Output the LPF(n).
Sample Input
12345
Sample Output
01213
Author
Wiskey
Source
HDU 2007-11 Programming Contest_WarmUp
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分解质因数:
每个合数都可以写成几个质数(即素数)相乘的形式,其中每个质数都是这个合数的因数,做这个合数的分解质因数。该题就是求输入n,对n进行分解质因数。在其所有的质因数中的最大的质因数在素数表中的位置。当然针对素数本身分解质因数可视为1*自身(注:分解质因数只针对合数)。
分解质因数算法:
/*分解质因数*/#include<iostream>using namespace std;int main(){ int n,n2; cin>>n; n2=n; cout<<n<<"="; for(int i=2;i<=n;i++) { for(;n2%i==0;) { n2=n2/i; cout<<i<<"*"; } if(n2==1) break; } cout<<"1"; return 0;}
测试结果:
结合分解质因数和筛法,AC的结果:
#include<stdio.h>#include<string.h>#define MAX 1000000using namespace std;int LPF[MAX];int main(){ //利用建立一张素数表,其index代表素数,其值LPF[index]代表位置信息 int i; int val; memset(LPF,0,sizeof(LPF)); LPF[2]=1; for(i=4;i<MAX;i+=2)//处理偶数 { LPF[i]=LPF[2]; } int cnt=2; for(i=3;i<MAX;i+=2)//处理奇数 { if(LPF[i])//即非0 continue; LPF[i]=cnt++; for(int j=(i<<1);j<MAX;j+=i) { LPF[j]=LPF[i]; } } //必须得用scanf与printf,涉及到效率问题,传送门http://bbs.csdn.net/topics/390774520 while(~scanf("%d",&val)) printf("%d\n",LPF[val]); return 0;}
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