time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
PolandBall is a young, clever Ball. He is interested in prime numbers. He has stated a following hypothesis: “There exists such a positive integer n that for each positive integer m number n·m + 1 is a prime number”.
Unfortunately, PolandBall is not experienced yet and doesn’t know that his hypothesis is incorrect. Could you prove it wrong? Write a program that finds a counterexample for any n.
Input
The only number in the input is n (1 ≤ n ≤ 1000) — number from the PolandBall’s hypothesis.
Output
Output such m that n·m + 1 is not a prime number. Your answer will be considered correct if you output any suitable m such that 1 ≤ m ≤ 103. It is guaranteed the the answer exists.
Examples
input
3
output
1
input
4
output
2
Note
A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.
For the first sample testcase, 3·1 + 1 = 4. We can output 1.
In the second sample testcase, 4·1 + 1 = 5. We cannot output 1 because 5 is prime. However, m = 2 is okay since 4·2 + 1 = 9, which is not a prime number.
【题目链接】:http://codeforces.com/contest/755/problem/A
【题解】
枚举m从1..1000就好;
(special way 注意到当m=n+2的时候,n*m+1=(n+1)^2,而当m=n-2的时候,n*m+1=(n-1)^2,所以对n分类讨论一下就可以了.)
代码给的是枚举的.
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 110;
int n;
bool is(int k)
{
int len = sqrt(k);
rep1(j,2,len)
if (k%j==0)
{
return false;
}
return true;
}
int main()
{
//freopen("F:\rush.txt","r",stdin);
rei(n);
rep1(m,1,1000)
{
int k = n*m+1;
//cout << k <<endl;
if (!is(k))
{
printf("%d
",m);
return 0;
}
}
return 0;
}