POj  3292 Semi-prime H-numbers

Semi-prime H-numbers

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 5400Accepted: 2209

Description

This problem is based on an exercise of David Hilbert, whopedagogically suggested that one study the theoryof 4n+1 numbers. Here,we do only a bit of that.

An H-number is a positive number whichis one more than a multiple of four: 1, 5, 9, 13, 17, 21,... arethe H-numbers. For this problem we pretendthat these arethe only numbers.The H-numbers are closed undermultiplication.

As with regular integers, we partitionthe H-numbers intounits, H-primes,and H-composites. 1 is the only unit.An H-number h is H-primeif it is not the unit, and is the product oftwo H-numbers in only one way: 1× h. The rest of the numbersare H-composite.

For examples, the first few H-compositesare: 5 × 5 = 25, 5 × 9 = 45, 5 × 13 = 65, 9 × 9 = 81, 5 × 17 =85.

Your task is to count the numberof H-semi-primes.An H-semi-prime isan H-number which is the product of exactlytwo H-primes. Thetwo H-primes may be equal or different. Inthe example above, all five numbersare H-semi-primes. 125 = 5 × 5 × 5 is notan H-semi-prime, because it's the productof 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 shouldnot be processed.

Output

For eachinputted H-number h,print a line stating h andthe number of H-semi-primes between 1and h inclusive, separatedby one space in the format shown in the sample.

Sample Input

21 857890

Sample Output

21 085 5789 62

Source

Waterloo LocalContest, 2006.9.30

唉，崩溃啊，差点儿超内存，32ms，9352K，

打表，记下所有“素数”，然后再打表记下个数，最后输出。

代码和测试数据如下：

#include<stdio.h>

intprm[1100008],hnum[1100008],a[1100008]={0},qun[1100008]={0};

__int64q;

intmain()

{

intm,n,i=0,j,k,num;

for(j=1;j<=1000001;j+=4)

{

hnum[i++]=j;

a[j]=1;

}m=i;k=0;

for(i=1;i<m;i++)

if(a[hnum[i]]==1)

{

prm[k++]=hnum[i];

for(j=i;j<m;j++)

{

q=hnum[i]*hnum[j];

if(q>1000001||(hnum[i]>1001&&hnum[j]>1001))break;

a[hnum[i]*hnum[j]]=2;

}

}

num=0;m=0;

for(i=0;i<k;i++)

for(j=i;j<k;j++)

{

q=hnum[i]*hnum[j];

if(q>1000001||(hnum[i]>1001&&hnum[j]>1001))break;

if(prm[i]*prm[j]<=1000001)

{

qun[prm[i]*prm[j]]=1;

a[m++]=prm[i]*prm[j];

}

}

for(i=1;i<=1000001;i++)qun[i]+=qun[i-1];

while(scanf("%d",&n),n)

{

printf("%d%d\n",n,qun[n]);

}

return0;

}

这是学长的代码，超简单，唉，学长就是学长啊！

#include<stdio.h>

#include<string.h>

#define N1000002

intflag[N],a[N];

intmain()

{

inti,j,n;

memset(flag,-1,sizeof(flag));

memset(a,0,sizeof(a));

for(i=5;i<1001;i+=4){

for(j=i;i*j<N;j+=4)

flag[i*j]=0;

}

for(i=5;i<1001;i+=4)

for(j=i;i*j<N&&flag[i];j+=4)

if(flag[j])

a[i*j]=1;

for(i=1;i<N;i++)a[i]+=a[i-1];

while(scanf("%d",&n),n)

printf("%d%d\n",n,a[n]);

return0;

}

测试数据：

1

10

45

452

100

1005

200

20013

1000

100082

2000

2000174

5000

5000468

4567

4567427

7891

7891764

100000

10000010394

500000

50000052755

1000001

1000001105753