Goldbach`s Conjecture(素筛水题)题解
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.
InputInput 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).
OutputFor 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 Input2
6
4
Sample OutputCase 1: 1
Case 2: 1
思路:
水题一只,叫你求n=a+b且a,b都是素数这样的ab的对数。只要素筛一下,就能做了。最好先储存一下10^7以内的素数,数答案时只要到n/2就行了(因为a<=b),不知道不弄会不会超时,可以试一下。
代码:
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
const int N=10000005; //18
const int MOD=1000;
using namespace std;
bool prime[N];
int p[N/10],pn;
void get_prime(){
pn=0;
memset(prime,false,sizeof(prime));
prime[0]=prime[1]=true;
for(long long i=2;i<N;i++){
if(!prime[i]){
p[pn++]=i;
for(long long j=i*i;j<N;j+=i){
prime[j]=true;
}
}
}
}
int main(){
get_prime();
int T,n,num=1,ans;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
ans=0;
for(int i=0;p[i]<=n/2;i++){
if(prime[n-p[i]]==false) ans++;
}
printf("Case %d: %d\n",num++,ans);
}
return 0;
}