| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7059 | Accepted: 3030 |
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 <algorithm>
#include <iostream> using namespace std; #define maxn 1000005 bool H[maxn];
int ans[maxn],ele[maxn];
int len = ; void init() { for(int i = ; i <= maxn - ; i++) {
H[i] = (i % == );
} for(int i = ; i * i <= maxn - ; i += ) {
if(!H[i]) continue;
for(int j = i; j * i <= maxn - ; j++) {
H[j * i] = ;
}
} for(int i = ; i <= maxn - ; i += ) {
if(H[i]) {
ele[len++] = i;
}
} for(int i = ; i < len && ele[i] * ele[i] <= maxn - ; i++) {
for(int j = i; j < len && ele[j] * ele[i] <= maxn - ; j++) {
if(ele[i] * ele[j] % == )
ans[ ele[i] * ele[j] ] = ;
}
} for(int i = ; i <= maxn - ; i++) {
ans[i] += ans[i - ];
}
} int main() {
// freopen("sw.in","r",stdin); init(); int x;
while(~scanf("%d",&x) && x) {
printf("%d %d\n",x,ans[x]);
} return ; }