UVA - 11105
Description
Problem A: Semi-prime H-numbers
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. 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 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 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. For each inputted H-number h, print a line stating h and the number of H-semi-primes between 1 and h inclusive, separated Sample input21 85 789 0 Output for sample input21 0 85 5 789 62 Don Reble Source Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: More Advanced Topics :: Problem Decomposition :: Two Root :: AOAPC II: Beginning Algorithm Contests (Second Edition) (Rujia Liu) :: Chapter 10. Maths :: Exercises Root :: AOAPC I: Beginning Algorithm Contests -- Training Guide (Rujia Liu) :: Chapter 2. Mathematics :: Number Theory :: Exercises: |
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先筛出所有的H素数，然后暴力搞。。。不知道大家求H半素数有没有什么更好的算法呢？虽然程序跑的很快，但是还是想知道是否有更快的。
#include<bits/stdc++.h>
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 5;
bool check[maxn];
int f[maxn];
void init(int n)
{
memset(check,0,sizeof check);
vector<int> res;
for(int i = 5; i <= n; i += 4) {
if(!check[i])res.push_back(i);
int sz = res.size();
for(int j = 0; j < sz; j++) {
ll t = (ll)i*res[j];
if(t>n)break;
check[t] = true;
if(i%res[j]==0)break;
}
}
memset(f,0,sizeof f);
int sz = res.size();
for(int i = 0; i < sz; i++) {
for(int j = i; j < sz; j++) {
ll t = (ll)res[i] * res[j];
if(t>n)break;
f[t] = 1;
}
}
for(int i = 1; i <= n; i++)f[i] += f[i-1];
}
int main()
{
int n;
init(maxn-5);
while(~scanf("%d",&n)&&n) {
printf("%d %d\n", n, f[n]);
}
return 0;
}