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Codeforces.

enter n and k, 2k <= n, output 2 * n, each of which isn't identical and <= 2 * n Thinking: first, we look at k = 0. When 2, 1, 4, 3...
A position, the first one.The formula.The value is unchanged, and the value of the next formula decreases 2, so the interpolation becomes 2, so we just exchange k logarithm, the value of the second equation is n
2k, the difference becomes 2 * k.
#include<iostream>
using namespace std;
int main(void)
{
 int n,k;
 cin>>n>>k;
 int flag=0;
 for(int i=1;i<=k;i++)
 {
 if(!flag) cout<<2*i-1<<""<<2*i;
 else cout<<""<<2*i-1<<""<<2*i;
 flag=1; 
 }
 for(int i=k+1;i<=n;i++)
 {
 if(!flag) cout<<2*i<<""<<2*i-1;
 else cout<<""<<2*i<<""<<2*i-1;
 flag=1;
 }
 return 0;
}






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