(Hdu Acm Step 2.1.3)Largest prime factor
2017-08-13 11:13
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题目:
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:给你一个数,让你求这个数最大质因子在素数中的位置,比如说2的最大质因子是2,位置就是1, 4和8的最大质因子也是2,位置还是1; 3的最大质因子是3,位置是2,6和9最大质因子也是3,位置是2。
AC代码:
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:给你一个数,让你求这个数最大质因子在素数中的位置,比如说2的最大质因子是2,位置就是1, 4和8的最大质因子也是2,位置还是1; 3的最大质因子是3,位置是2,6和9最大质因子也是3,位置是2。
AC代码:
#include<stdio.h> #include<string.h> #define max 1000001 int a[max]; /*同一个数的质因子位置可能被多次复写。 但是由于是从小到大遍历, 这就保证了最后一次写入的是该数的最大质因子的位置 比如说第一次i=2时,a[2],a[4],a[6],a[8],a[10],a[12]最大质因子的位置是1; 在第二次 i=3时,a[3],a[6],a[9],a[12],最大质因子的位置等于是2, 这时a[6]和a[12]会将第一次的值覆盖 */ int main() { int n; int i,j,count = 1; memset(a, 0, sizeof(a)); a[1] = 0; for(i = 2; i < max; i++) { if(a[i] == 0) { for(j = i; j < max; j += i) a[j] = count; count++; } } while(scanf("%d", &n) != EOF) { printf("%d\n", a ); } return 0; }
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