Codeforces Round #324 (Div. 2) D. Dima and Lisa 数论 三素数定理

D. Dima and Lisa

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Dima loves representing an odd number as the sum of multiple primes, and Lisa loves it when there are at most three primes. Help them to represent the given number as the sum of at most than three primes.

More formally, you are given an odd numer n. Find a set of numbers pi (1 ≤ i ≤ k), such that

1 ≤ k ≤ 3

pi is a prime

The numbers pi do not necessarily have to be distinct. It is guaranteed that at least one possible solution exists.

Input

The single line contains an odd number n (3 ≤ n < 109).

Output

In the first line print k (1 ≤ k ≤ 3), showing how many numbers are in the representation you found.

In the second line print numbers pi in any order. If there are multiple possible solutions, you can print any of them.

Sample test(s)

input

27

output

3

5 11 11

Note

A prime is an integer strictly larger than one that is divisible only by one and by itself.

题意，要求一个奇数分解成小于3个质数和

三素数定理

根据哥德巴赫猜想，任意大的偶数都可以分解两个质数和，那么，把奇数，先减去一个最接近的质数，就得到了一个偶数，枚举这两个质数就可以了。由于两个质数间距是非常小的，小于800，所以很快就可以求解。

总复杂度不超过O(sqrt(n) * 800);

#define N 205
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n;
bool IsPrime(int n)
{
int n1 = sqrt(n );
for (int i=2;i<=n1;++i)
{
if (n%i==0)
{
return false;
}
}
return true;
}
int main()
{
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
while(S(n)!=EOF)
{
if(n == 3){
printf("1\n");
printf("3\n");
continue;
}
if(n == 5){
printf("2\n");
printf("2 3\n");
continue;
}
if(n == 7){
printf("3\n");
printf("2 2 3\n");
continue;
}
if(n == 9){
printf("3\n");
printf("2 2 5\n");
continue;
}
if(n == 11){
printf("3\n");
printf("2 2 7\n");
continue;
}
int s1,s2,s3;
For(i,2,800){
if(n - i < 0) break;
if(IsPrime(n - i)){
s1 = n - i;
break;
}
}
n = n - s1;
//printf("nn %d\n",n);
if(IsPrime(n)){
printf("2\n");
printf("%d %d\n",s1,n);
}
int n2 = n / 2 + 1;
For(i,2,n2){
if(IsPrime(i)&& IsPrime(n - i)){
printf("3\n");
printf("%d %d %d\n",i,n-i,s1);
break;
}
}
}
//fclose(stdin);
//fclose(stdout);
return 0;
}