P1304 godebach conjecture, p1304 godebach Conjecture

P1304 godebach conjecture, p1304 godebach Conjecture
Description

Input N (N <= 10000), verify 4 ~ N: whether all the even numbers comply with the godebach conjecture.

(N is an even number ).

If a number is used, for example, 10, the first addition number is output, which is the smallest solution compared to other solutions. For example, 10 = 3 + 7 = 5 + 5, 10 = 5 + 5 is the incorrect answer.

Input/Output Format

Input Format:

 

First line N

 

Output Format:

 

4 = 2 + 2 6 = 3 + 3 ...... N = x + y

 

Input and Output sample input sample #1:

10

Output sample #1:

4 = 2 + 26 = 3 + 38 = 3 + 510 = 3 + 7

First screen one side of prime number,
Then the brute-force enumeration is good.

 1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<queue> 6 using namespace std; 7 const int MAXN=10001; 8 const int maxn=0x7fffff; 9 void read(int &n)10 {11     char c='+';int x=0;bool flag=0;12     while(c<'0'||c>'9')13     {c=getchar();if(c=='-')flag=1;}14     while(c>='0'&&c<='9')15     {x=x*10+(c-48);c=getchar();}16     flag==1?n=-x:n=x;17 }18 int n;19 int vis[MAXN];20 int main()21 {22     read(n);23     vis[1]=1;24     for(int i=2;i<=n;i++)25         if(!vis[i])26         for(int j=i*i;j<=n;j+=i)27             vis[j]=1;28     for(int i=4;i<=n;i+=2)29         for(int j=2;j<=n;j++)30             if(vis[j]==0&&vis[i-j]==0)31             {32                 printf("%d=%d+%d\n",i,j,i-j);break;33             }34                 35     return 0;36 }