poj3292 Semi-prime H-numbers
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 9873 | Accepted: 4404 |
Description
This problem is based on an exercise of David Hilbert, who pedagogically suggested that one study the theory of 4n+1 numbers. Here, we do only a bit of that.
An H-number is a positive number which is one more than a multiple of four: 1, 5, 9, 13, 17, 21,... are the H-numbers. For this problem we pretend that these are the only numbers. The H-numbers are closed under multiplication.
As with regular integers, we partition the H-numbers into units, H-primes, and H-composites. 1 is the only unit. An H-number h is H-prime if it is not the unit, and is the product of two H-numbers in only one way: 1 × h. The rest of the numbers are H-composite.
For examples, the first few H-composites are: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 = 85.
Your task is to count the number of H-semi-primes. An H-semi-prime is an H-number which is the product of exactly two H-primes. The two H-primes may be equal or different. In the example above, all five numbers are H-semi-primes. 125 = 5 × 5 × 5 is not an H-semi-prime, because it's the product of three H-primes.
Input
Each line of input contains an H-number ≤ 1,000,001. The last line of input contains 0 and this line should not be processed.
Output
For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated by one space in the format shown in the sample.
Sample Input
21 85 789 0
Sample Output
21 0 85 5 789 62
Source
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <cmath> using namespace std; const int maxn = 1000001; int flag[maxn],sum[maxn],h; void init() { for (int i = 5; i <= maxn; i += 4) for (int j = 5; j <= maxn; j += 4) { int t = i * j; if (t > maxn) break; if (!flag[i] && !flag[j]) flag[t] = 1; else flag[t] = -1; } int cnt = 0; for (int i = 1; i <= maxn; i++) { if (flag[i] == 1) cnt++; sum[i] = cnt; } } int main() { init(); while (scanf("%d",&h) && h != 0) printf("%d %d\n",h,sum[h]); return 0; }