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
Hint
- An integer is said to be prime, if it is divisible by exactly two different integers. First few primes are 2, 3, 5, 7, 11, 13, …
解题报告
先得对107 内的素数打表只有664580个,如此数据量级就减小了100倍
判断素数a+b=n,只要取素数a,在判断 n-a 是否为素数即可复杂度o(φ(n))
#include<stdio.h>
#include<algorithm>
#define MAX_N 10000020
#define P_N 664580
using namespace std;
bool vis[MAX_N];
int p[P_N];
void init(){
int cnt=0;
for(int i=2;i*i<=MAX_N;i++)
if(!vis[i])
for(int k=i*i;k<MAX_N;k+=i)
vis[k]=true;
for(int i=2;i<MAX_N;i++)
if(!vis[i])
p[cnt++]=i;
}
int main()
{
init();int T;
scanf("%d",&T);
for(int t=1;t<=T;t++){
int n;
scanf("%d",&n);
int ans=0;
for(int i=0;p[i]<=n/2;i++){
if(!vis[n-p[i]]) ans++;
}
printf("Case %d: %d\n",t,ans);
}
return 0;
}