素数筛 LightOJ - 1259 Goldbach`s Conjecture
Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:
Every even integer, greater than 2, can be expressed as the sum of two primes [1].
Now your task is to check whether this conjecture holds for integers up to 107.
Input
Input starts with an integer T (≤ 300), denoting the number of test cases.
Each case starts with a line containing an integer n (4 ≤ n ≤ 107, n is even).
Output
For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of (a, b) where
1) Both a and b are prime
2) a + b = n
3) a ≤ b
Sample Input
2
6
4
Sample Output
Case 1: 1
Case 2: 1
题意就是找一个偶数可以分解为几对素数之和,数据有些大,需要用筛法求素数。
#include <iostream>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <stdio.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int maxn=1e7;
bool vis[maxn];
int pri[maxn/10];
void prime()
{
memset(vis,0,sizeof(vis));
int num=0;
for(int i=2;i<maxn;i++)
{
if(!vis[i]) //原理就是每次标记所有约数有i的数,然后遍历所有没有被标记过的数
{
pri[++num]=i;
for(int j=i*2;j<=maxn;j+=i)
vis[j]=1;
}
}
}
int main()
{
prime();
int t,n;
cin>>t;
for(int cas=1;cas<=t;cas++)
{
cin>>n;
int ans=0,tmp=n/2;
for(int i=1;pri[i]<=tmp;i++)
{
if(vis[n-pri[i]]==0)
{
ans++;
}
}
cout<<"Case "<<cas<<": "<<ans<<endl;
}
return 0;
}