Largest prime factor
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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
这题因为数字比较庞大,直接暴力算会超时,所以需要打表,但不是打素数表,打素数表在进行判断也会超时。需要打每个数的最大素数因子表,对于我来讲,认为技巧性挺高,当然,难了不会,会了不难。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#define maxn 1000000
using namespace std;
int visited[maxn],n;
void fact(){
int num=0;
for(int i=2;i<maxn;i++){//从最小素数2开始
if(visited[i]==0) //num++;如果这个数没有被之前的素数刷新过,那么这个数就一定是素数。
else continue;//只有是素数,才能进行刷新操作
for(int j=i;j<maxn;j+=i){//如果是素数的倍数,那么将被刷新。
visited[j]=num;
}
}
}
int main(){
fact();
while(scanf("%d",&n)!=EOF){
printf("%d\n",visited[n]);
}
return 0;
}
#include<cstdio>
#include<cmath>
#include<cstring>
#define maxn 1000000
using namespace std;
int visited[maxn],n;
void fact(){
int num=0;
for(int i=2;i<maxn;i++){//从最小素数2开始
if(visited[i]==0) //num++;如果这个数没有被之前的素数刷新过,那么这个数就一定是素数。
else continue;//只有是素数,才能进行刷新操作
for(int j=i;j<maxn;j+=i){//如果是素数的倍数,那么将被刷新。
visited[j]=num;
}
}
}
int main(){
fact();
while(scanf("%d",&n)!=EOF){
printf("%d\n",visited[n]);
}
return 0;
}
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