AtCoder Beginner Contest 096 D - Five, Five Everywhere



Problem Statement

Print a sequence a₁,a₂,…,a_N whose length is N that satisfies the following conditions:

• a_i (1≤i≤N) is a prime number at most 55 555.
• The values of a₁,a₂,…,a_N are all different.
• In every choice of five different integers from a₁,a₂,…,a_N, the sum of those integers is a composite number.

If there are multiple such sequences, printing any of them is accepted.

Notes

An integer N not less than 2 is called a prime number if it cannot be divided evenly by any integers except 1 and N, and called a composite number otherwise.

Constraints

• N is an integer between 5 and 55 (inclusive).

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Input

Input is given from Standard Input in the following format:

N

Output

Print N numbers a₁,a₂,a₃,…,a_N in a line, with spaces in between.

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Sample Input 1

Copy

5

Sample Output 1

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3 5 7 11 31

Let us see if this output actually satisfies the conditions.
First, 3, 5, 7, 11 and 31 are all different, and all of them are prime numbers.
The only way to choose five among them is to choose all of them, whose sum is a₁+a₂+a₃+a₄+a₅=57, which is a composite number.
There are also other possible outputs, such as 2 3 5 7 13, 11 13 17 19 31 and 7 11 5 31 3.

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Sample Input 2

Copy

6

Sample Output 2

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2 3 5 7 11 13

• 2, 3, 5, 7, 11, 13 are all different prime numbers.
• 2+3+5+7+11=28 is a composite number.
• 2+3+5+7+13=30 is a composite number.
• 2+3+5+11+13=34 is a composite number.
• 2+3+7+11+13=36 is a composite number.
• 2+5+7+11+13=38 is a composite number.
• 3+5+7+11+13=39 is a composite number.

Thus, the sequence 2 3 5 7 11 13 satisfies the conditions.

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Sample Input 3

Copy

8

Sample Output 3

Copy

2 5 7 13 19 37 67 79

题意：给出长度为n的素数序列，随机从其中抽出5个数相加的和都为合数，求这样的序列

思路：因为是5个素数相加的和为合数，所以只要5个数中的数字个位相同即可

#include <bits/stdc++.h>
using namespace std;
int num[20000];
int pr[100];

int is_prime(int n)
{
    int m=sqrt(n+0.5);
    for(int i=2; i<=m; i++)
        if(n%i==0)
            return 0;
    return 1;
} 

int main()
{
    int cnt=0;
    for(int i=3; i<55555; i++)  //判断素数
        if(is_prime(i))
            num[cnt++]=i;
    int k=0;
    for(int i=0; i<cnt; i++)  //记录位数
    {
        if((num[i]-1)%5==0)
            pr[k++]=num[i];
        if(k==60)
            break;
    }
    int n;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        if(i!=0)
            cout << ' ';
        cout << pr[i];
    }
    cout << endl;
    return 0;
}