hdu 2136(质数筛选+整数分解)

Largest prime factor

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9993    Accepted Submission(s): 3528

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.

 

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

 

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const int N = ;
bool p[N]; ///为false代表是素数
int idx[N];
void init(){
    memset(p,false,sizeof(p));
    int id = ;
    for(int i=;i<N;i++){
        if(!p[i]){
            idx[i] = id++;
            for(LL j=(LL)i*i;j<N;j+=i){
                p[j] = true;
            }
        }
    }
}
int getMax(int n){
    int Max = -;
    for(int i=;i*i<=n;i++){
        if(n%i==){
            while(n%i==){
                n/=i;
            }
            Max = max(i,Max);
        }
    }
    if(n>) Max = max(Max,n);
    return Max;
}
int main()
{
    init();
    int n;
    while(scanf("%d",&n)!=EOF){
        if(n==) printf("0\n");
        else{
            int Max = getMax(n);
            printf("%d\n",idx[Max]);
        }
    }
}

O(n)的素数筛

/*求第n个质数*/
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
typedef long long LL;
const int N = ;
int n,m;
int p[];//存储素数
bool a[N];

//O(n) 素数筛
void init() {
    memset(a,false,sizeof(a));//初始全部为素数
    int num=;
    for(int i=;i<N;++i) {
        if(!a[i]) p[num++]=i;
        for(int j=;(j<num&&i*p[j]<N);++j) {
            a[i*p[j]]=;
            if(i%p[j] == ) break;
        }
    }
}

int main(){
    init();
    int n;
    int k = ;
    while(scanf("%d",&n)!=EOF){
       if(n==) break;
       printf("Case %d: %d\n",++k,p[n-]);
    }
}