HDU 2136 Largest prime factor(筛选法)
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Largest prime factor
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4026 Accepted Submission(s): 1415
Problem 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
1
2
3
4
5
Sample Output
0
1
2
1
3
Author
Wiskey
Source
HDU 2007-11 Programming Contest_WarmUp
Recommend
威士忌
分析:用筛选法(分解素数技巧),求N的最大素数因子在素数表的位置
cout与printf 输出效率是不一样的
//cout<<a[n]<<endl;
printf("%d\n",a[n]);
printf("%d\n",a[n]);
#include<iostream>using namespace std;int a[1000000];int main(){ int k=1;//k初始化为1,位置 memset(a,0,sizeof(a)); /*筛选位置*/ for(int i=2;i<1000000;i++)//要到N,而不是sqrt(N) { if(!a[i]) { a[i]=k;//素数i的位置 for(int j=i+i;j<1000000;j+=i) a[j]=k; //构造出j的暂时最大素数因子的位置 k++; } } int n; while(~scanf("%d",&n)) { printf("%d\n",a[n]); } return 0;}
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