Codeforces round 275 (Div. 2) C – diverse permutation (construction) [easy to understand]

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Topic link ：Codeforces Round #275 (Div. 2) C – Diverse Permutation

The question ： A string of 1~n. Find the number of absolute values of the difference between two adjacent terms in a sequence （ The number of different absolute values ） by k individual . Find sequence 、

Ideas ：1~k+1. Construction sequence front segment , Then directly output the remaining number . The previous structure can be based on , The absolute value of the difference between the two terms is 1~k structure .

AC Code ：

#include <stdio.h>
#include <string.h>
int ans[200010];
bool vis[100010];
int n,mark;
int iabs(int a)
{
    if(a<0) return -a;
    return a;
}
int main()
{
    int i,cnt,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        int x,y;
        memset(vis,0,sizeof vis);
        ans[0]=1;
        x=1,y=k+1;
        cnt=k;
        for(i=1; i<=k; i++,cnt--)
        {
            int temp=ans[i-1]+cnt;
            if(temp>k+1)
                temp=ans[i-1]-cnt;
            else if(vis[temp])
                temp=ans[i-1]-cnt;
            ans[i]=temp;
            vis[temp]=true;
        }
        for(i=k+1; i<n; i++)
            ans[i]=i+1;
        for(i=0; i<n-1; i++)
            printf("%d ",ans[i]);
        printf("%d\n",ans[i]);
    }
    return 0;
}

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