Let’s call a positive integer composite if it has at least one divisor other than 1 and itself. For example:
the following numbers are composite: 1024, 4, 6, 9;
the following numbers are not composite: 13, 1, 2, 3, 37.
You 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.
Examples
inputCopy
1
outputCopy
9 8
inputCopy
512
outputCopy
4608 4096
题意:找出两个合数a,b使得a-b=n,给出n
解析:
第一种做法:a-b=n,那么a=b+n,我们枚举b check(b)&&check(b+n),循环次数的一定很少因为素数很少。
第二种做法:a-b=n,那么a=3n,b=2n==3n-2n=n。对于1直接输出样例,对于其他的输出a3,b2;
第一种:
#include<bits/stdc++.h>
using namespace std;
int n,a,b;
int check(int x)
{
int i;
for( i=2;i<=sqrt(x)&&x%i;i++);
if(i>sqrt(x)) return 0;
return 1;
}
int main()
{
cin>>n;
for(int i=4;;i++)
{
if(check(i)&&check(i+n))
{
b=i;
a=i+n;
break;
}
}
cout<<a<<" "<<b<<endl;
}
第二种:
#include<bits/stdc++.h>
using namespace std;
int n,a,b;
int main()
{
cin>>n;
if(n==1)
{
cout<<9<<" "<<8<<endl;
}
else cout<<n*3<<" "<<n*2<<endl;
}