CCNU第五周数学练习的一道水题-白红宇

CCNU第五周数学练习的一道水题

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发布时间：2019-05-27

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E - Semi-prime H-numbers

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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 857890

Sample Output

21 085 5789 62题目意思就是诸如4N+1的数我们把它叫做H-NUMBER数 然后我们现在要来找一个范围内的H-semi-primes数量有多少 而H-semi-primes这个数呢就是它能且只能由两个H-PRIMES数相乘得到（由多组H-PRIMES数相乘得到没有关系 例如441=21*21=9*49但是它也属于H-semi-primes范畴），而H-PRIMES数就是只能由1和H-NUMBER数相乘得到的数要注意题目中给的一个条件The H-numbers are closed under multiplication.说明了这个乘法运算是闭包的也就是说（4*N+1）*(4*M+1)=4*K+1 其中N M K均为大于等于1的正整数所以做这道题我们只需要对满足条件的数进行暴力的筛选好了，然后全部筛选出来后再来统计下满足条件的个数就可以了。很简单。。没有数论的知识在里面，但是我不知道如果这道题数字再继续变大的的话暴力筛选会不会被TLE下面贴上我的代码。。。 #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define MAXN 10000001int H[MAXN+1];void choose(){    int i,j;    memset(H,0,sizeof(H));    for(i=5;i<=MAXN;i+=4)    {        for(j=5;j<=MAXN;j+=4)        {            if(i*j>MAXN)break;            if(H[i]==0&&H[j]==0)                H[i*j]=1;            else                H[i*j]=-1;        }    }int count=0;for(i=1;i<=MAXN;i++)    {        if(H[i]==1)        count++;        H[i]=count;    }}int main(){    int h;    choose();    while(scanf("%d",&h)!=EOF)    {        if(h==0)            return 0;        cout<
          
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