poj3292——筛选法
poj3292——筛选法
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 7772 | Accepted: 3360 |
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 twoH-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
题意:不解释了,自己看吧
思路:先筛H素数,再筛Hsemi数,最后计数打表
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> using namespace std; const int maxn=1000100; const int INF=(1<<28); int h; bool Hprime[maxn]; bool Hsemi[maxn]; int ans[maxn]; void play_list() { memset(Hprime,1,sizeof(Hprime)); for(int i=5;i<maxn;i+=4){///筛选H素数 for(int j=5;j<maxn;j+=4){ ///把j的初值改成i就WA。。。因为下面的break使提前退出了? int mul=i*j; if(mul>=maxn||mul<=0) break; Hprime[mul]=0; } } memset(Hsemi,0,sizeof(Hsemi)); for(int i=5;i<maxn;i+=4){///筛选Hsemi数 for(int j=5;j<maxn;j+=4){ ///这里也是,把j的初值改成i就WA。。。 int mul=i*j; if(mul>=maxn||mul<=0) break; if(Hprime[i]&&Hprime[j]) Hsemi[mul]=1; } } memset(ans,0,sizeof(ans)); for(int i=1;i<maxn;i++){///打表计数 if(Hsemi[i]) ans[i]=ans[i-1]+1; else ans[i]=ans[i-1]; } } void debug() { for(int i=100000;i<=100100;i++){ cout<<i<<" "<<ans[i]<<endl; } } int main() { play_list(); //debug(); while(cin>>h,h){ cout<<h<<" "<<ans[h]<<endl; } return 0; }