Goldbach`s Conjecture （数论+思维）

 Goldbach`s Conjecture

1259 - Goldbach`s Conjecture

PDF (English)

Statistics

Forum

Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
2 6 4	Case 1: 1 Case 2: 1

Note

1.      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, ...

 

wa了n发。。。。。。稀里糊涂就过了。。。。

#include
#define ll long long
using namespace std;
const int maxn = 1e6+7;

int prim[5*maxn], cnt = 0;
bool vis[10*maxn];

void getprim()                                  //素数打表；
{
    memset(vis, 0, sizeof(vis));
    for(int i = 2; i*i < 10*maxn; i++)
    {
        if(!vis[i]) {
            for(int j = i*i; j < 10*maxn; j+= i)
                vis[j] = true;
        }
    }
    for(int i = 2; i < 5*maxn; i++)
        if(!vis[i]) prim[cnt++] = i;
}

int Find(int n)
{
    int sum = 0;
    for(int i = 0; i < cnt; i++)
    {
        if(prim[i] > n/2) break;
        else if(!vis[n - prim[i]]) sum++;         //注意素数表开到1e7+7；
    }
    return sum;
}

int main()
{
    int t,n, cas = 0;
    getprim();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case %d: %d\n", ++cas, Find(n));
    }
}