LightOJ 1259 Goldbach`s Conjecture（数论）

Goldbach`s Conjecture

Description

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

解题思路：

4 ≤ n ≤ 10的7次方，数据稍微有点大，因此用素数筛选法时定义数组要用bool类型，因为bool只占1个字节，而int占4个字节。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

bool vis[10000005];//bool只占1个字节，int占4个字节
vector<int> v;

void get_prime(){
memset(vis,false,sizeof(vis));
vis[1] = true;
for(int i = 2; i <= 10000000; i++){
int t = 10000000/i;
for(int j = 2; j <= t; j++)
vis[i*j] = true;
}
for(int i = 2; i <= 10000000; i++)
if(!vis[i])
v.push_back(i);
}

int main(){
get_prime();
int T,t = 1;
scanf("%d",&T);
while(T--){
int n,ans = 0;
scanf("%d",&n);
for(int i = 0; v[i] <= n/2; i++){
int tmp = n-v[i];
if(!vis[tmp])
ans++;
}
printf("Case %d: %d\n",t++,ans);
}
return 0;
}