【HDU 2136】Largest prime factor（素数）



Largest prime factor

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11955    Accepted Submission(s): 4240

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

题目大意：n的最大质因数是第几个质数

思路：筛法打表。最初先找出每个n的质因数，结果T了，后来发现打表时可以将每个数的最大质因数记录下来，数组p表示下标的最大质因数，不断更新p，最终结果就是最大的质因数。

#include<bits/stdc++.h>#define manx 1000006using namespace std;int p[manx],a[manx],num[manx];void Prime(){    int cot=1;    memset(a, 1, sizeof(a));    for (int i=2; i<=manx; i++){        if(a[i]){                num[i]=cot++;    //质数i的位置                p[i]=i;            for (int j=i+i; j<=manx; j+=i){                p[j]=i;       //j的最大质因数                a[j]=0;            }        }    }}//int fun(int n)   T了//{//    for (int i=n; i>=2; i--)//        if(a[i]) return i;//}void solve(){    int n;    while(~scanf("%d",&n)){        if(n==1) {            printf("0\n");            continue;        }        printf("%d\n",num[p[n]]);    }}int main(){    Prime();    solve();    return 0;}

暑期集训又见了这题，总觉得似曾相识啊，原来真的写过，这次简洁了一点

#include <iostream>#include <cstdio>#define maxn 1000006using namespace std;int f[maxn];void Prime(){    int cot = 1;    for (int i = 2; i < maxn; i++){        if(!f[i]){            f[i] = cot++;            for (int j = i; j < maxn; j += i)                f[j] = f[i];        }    }}int main(){    Prime();    int n;    while(~scanf("%d",&n))        printf("%d\n",f[n]);    return 0;}