poj3292 Semi-prime H-numbers

Semi-prime H-numbers

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8845		Accepted: 3896

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

这个题一直哇！哇!的QAQ。
题意：在这里我们认为只有4n+1的数就是5...9...13....4n+1，然后我们规定有3类数：

1.H-prim 顾名思义，就是在这个系统里面的质数，因子没有别的4n+1的数。注意是在这个4n+1系统里的而不是整个实数哦～

2.H-semi-prim两个素数的乘积，注意！是准确的两个素数的乘积，像125可能有3个就不对了。

3.其余的那些数就是H-composites～了。

思路：类似素数筛的方法，把1-1000001的素数筛出来，然后依次枚举两个的乘积，用一个数组加标记，最后累加就可以辣～

其余的也没什么好说的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=1000000+10;
bool prim[MAXN];
int primnum[MAXN];
int ans[MAXN];
int ha[MAXN];
int hcnt;
void get_prim()
{
int i,j;
hcnt=0;
for(i=5; i<=1000001; i+=4)
{
if(prim[i])continue;
primnum[hcnt++]=i;
for(j=i; j<=1000001/i; j+=4)prim[i*j]=1;
}
}
void get_ans()
{
int i,j;
for(i=0; i<hcnt; ++i)
{
for(j=i; j<hcnt; ++j)
{
long long x=(long long)primnum[i]*primnum[j];
if(x>1000001)break;
ha[x]++;
}
}
int cnt=0;
for(i=5;i<=1000001;i+=4)
{
if(ha[i])ans[i]=++cnt;
else ans[i]=cnt;
}
}

int main()
{
int n;

get_prim();
get_ans();
while(~scanf("%d",&n)&&n)
{
printf("%d %d\n",n,ans
);
}
return 0;
}

还有一种做法就是，整体来筛选。（看起来就像是直接暴力枚举两个数的乘积）

因为这个系统无非就是3种数：素数，我们要找的数，和那个其余数。

不难想，任何数都可以最终分解成n个素数乘积的形式。所以我们可以这样做：

起始全标记为0（默认为素数），依次枚举两个的乘积，如果是两个素数的乘积（我们要找的数）就标记上1，否则就标记上个别的（其余数）。

最后累加1的就可以了～

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN=1000000+10;
int prim[MAXN];
int ans[MAXN];
int ha[MAXN];
int hcnt;
void get_prim()
{
int i,j;
hcnt=0;
for(i=5; i<=sqrt(1000001); i+=4)//其实思想依旧是欧拉筛的思想
{
for(j=i; j<=1000001/i; j+=4)
{
if(!prim[i]&&!prim[j])prim[i*j]=1;
else prim[i*j]=-1;
}
}
}
void get_ans()
{
int i;
int cnt=0;
for(i=5; i<=1000001; i+=4)
{
if(prim[i]==1)ans[i]=++cnt;
else ans[i]=cnt;
}
/* for(i=5; i<=10001; ++i)
{
if(prim[i]==1)
printf("%d ",i);
}*/

}

int main()
{
int n;

get_prim();
get_ans();
while(~scanf("%d",&n)&&n)
{
printf("%d %d\n",n,ans
);
}
return 0;
}