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.
InputEach line will contain one integer n(0 < n < 1000000).
OutputOutput the LPF(n).
Sample Input
1 2 3 4 5
Sample Output
0 1 2 1 3
对x分解质因数,问最大的质因子是第几大质数
瞎暴力就好
1 #include<cstdio> 2 #include<algorithm> 3 #define LL long long 4 using namespace std; 5 inline LL read() 6 { 7 LL x=0,f=1;char ch=getchar(); 8 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 9 while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 10 return x*f; 11 } 12 int mk[1000010]; 13 int pp[300010],len; 14 int rnk[1000010]; 15 int n; 16 inline void getp() 17 { 18 for (int i=2;i<=1000000;i++) 19 { 20 if (!mk[i]) 21 { 22 for (int j=2*i;j<=1000000;j+=i)mk[j]=1; 23 pp[++len]=i; 24 rnk[i]=len; 25 } 26 } 27 } 28 int main() 29 { 30 getp(); 31 while (~scanf("%d",&n)) 32 { 33 if (!mk[n]){printf("%d ",rnk[n]);continue;} 34 int mx=0; 35 for (int i=1;i<=len;i++) 36 { 37 if (pp[i]*pp[i]>n)break; 38 if (n%pp[i]==0)mx=i; 39 while (n%pp[i]==0)n/=pp[i]; 40 } 41 if (n!=1)mx=rnk[n]; 42 printf("%d ",mx); 43 } 44 }