A. Equation CodeForces - 1269A

一、内容

Let's call a positive integer composite if it has at least one divisor other than 1and itself. For example:  the following numbers are composite: 1024, 4, 6, 9;the following numbers are not composite: 13, 1, 2, 3, 37You are given a positive integer n. Find two composite integers a,b such that a−b=n.It can be proven that solution always exists.

Input
The input contains one integer n(1≤n≤107): the given integer.
Output

Print two composite integers a,b(2≤a,b≤109,a−b=n).It can be proven, that solution always exists.
If there are several possible solutions, you can print any.

Input

1

Output

9 8

二、思路

• 线性筛法暴力处理一下

三、代码

#include <cstdio>
const int N = 1e7 + 5000;
int n, cnt, pri[N + 5];
bool v[N + 5];
void init() {
	v[1] = true;
	for (int i = 2; i <= N; i++) {
		if (!v[i]) pri[cnt++] = i; 
		for (int j = 0; pri[j] <= N / i; j++) {
			v[i * pri[j]] = true;
			if (i % pri[j] == 0) break;
		}
	} 
}
int main() {
	scanf("%d", &n);
	init();
	for (int i = 4; ; i++) {
		if (v[i] && v[i + n]) {
			printf("%d %d", i + n, i); 
			break;
		}
	}
	return 0;
}