Poj 3292(篩法變形)
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 6750 | Accepted: 2868 |
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
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
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
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
就是在它定義的數上做篩法就可以了,需要真正理解這個演算法。#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
using namespace std;
const int maxn = 1000000 + 5;
const int INF = 2000000000;
typedef pair<int, int> P;
typedef long long LL;
vector<LL> h;
int p[maxn];
int sum[maxn];
void pre(){
h.clear();
for(int i = 1;i < maxn;i+=4){
h.push_back(i);
}
memset(p,0,sizeof(p));
for(int i = 1;i < h.size();i++){
if(p[h[i]] == 0){
for(int j = i;h[i]*h[j] < maxn;j++){
if(p[h[j]] == 0) p[h[i]*h[j]] = 1;
else p[h[i]*h[j]] = 2;
}
}
}
sum[1] = 0;
for(int i = 2;i < maxn;i++){
sum [i] = sum[i-1];
if(p[i] == 1) sum[i]++;
}
}
int main(){
pre();
int n;
while(scanf("%d",&n)){
if(n == 0) break;
printf("%d %d\n",n,sum[n]);
}
return 0;
}