poj3292
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 8677 | Accepted: 3793 |
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
大致题意:
一个H-number是所有的模四余一的数。
如果一个H-number是H-primes 当且仅当它的因数只有1和它本身(除1外)。
一个H-number是H-semi-prime当且仅当它只由两个H-primes的乘积表示。
H-number剩下其他的数均为H-composite。
给你一个数h,问1到h有多少个H-semi-prime数。
解题思路:
感觉跟同余模扯不上关系。。。
筛法打表,再直接输出。。。水题。。。
ps:请用G++提交
#include<iostream>
using namespace std;
const int N=;
int h,a[N+];
int go(){
for(int i=;i<=N;i+=){
for(int j=;j<=N;j+=){
int tmp=i*j;
if(tmp>N) break;
if(!a[i]&&!a[j])//i与j均为H-prime
a[tmp]=; //tmp为H-semi-primes
else
a[tmp]=-;//tmp为H-composite
}
}
int p=; //H-prime计数器
for(int i=;i<=N;i++){
if(a[i]==) p++;
a[i]=p; //从1到i有p个H-semi-primes
}
}
int main(){
go();
while(cin>>h){
if(!h) break;
cout<<h<<' '<<a[h]<<endl;
}
return ;
}
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