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http://codeforces.com/contest/482/problem/A
Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We‘ll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
3 2
3 1
5 2
1 3 2
1 2 3
1 3 2 4 5
By |x| we denote the absolute value of number x.
题意
从1-n的数，让你选择一些数来构造，要求每个相邻的数之间的绝对值之差有k种
题解：
按照1,n,2,n-1,3,n-2……这样子构造k-1组，然后把剩下没有遍历的都输出一下就好了
然后还有一种构造是n,1,2,n-1,3,n-2 这样子构造k-1组
这两种构造方法相差1，分别是对应k为奇数和偶数的情况
代码：
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 200001 #define mod 10007 #define eps 1e-9 //const int inf=0x7fffffff; //无限大 const int inf=0x3f3f3f3f; /* */ //************************************************************************************** inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } int main() { int flag[maxn]; int n,k; int ans=0; cin>>n>>k; if(k%2==0) { for(int i=0;i<k-1;i++) { if(i%2) { cout<<ans<<" "; flag[ans]=1; } else { cout<<n-ans<<" "; flag[n-ans]=1; ans++; } } } else { for(int i=0;i<k-1;i++) { if(i%2==0) { cout<<ans+1<<" "; flag[ans+1]=1; } else { cout<<n-ans<<" "; flag[n-ans]=1; ans++; } } } for(int i=1;i<=n;i++) { if(flag[i]==0) cout<<i<<" "; } return 0; }
Codeforces Round #275 (Div. 1)A. Diverse Permutation 构造
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原文地址：http://www.cnblogs.com/qscqesze/p/4385652.html
