B. Levko and Permutation 简单构造
Levko loves permutations very much. A permutation of length n is a sequence of distinct positive integers, each is at most n.
Let’s assume that value gcd(a, b) shows the greatest common divisor of numbers a and b. Levko assumes that element pi of permutation p1, p2, ... , pn is good if gcd(i, pi) > 1. Levko considers a permutation beautiful, if it has exactly k good elements. Unfortunately, he doesn’t know any beautiful permutation. Your task is to help him to find at least one of them.
The single line contains two integers n and k (1 ≤ n ≤ 105, 0 ≤ k ≤ n).
In a single line print either any beautiful permutation or -1, if such permutation doesn’t exist.
If there are multiple suitable permutations, you are allowed to print any of them.
4 2
2 4 3 1
1 1
-1
In the first sample elements 4 and 3 are good because gcd(2, 4) = 2 > 1 and gcd(3, 3) = 3 > 1. Elements 2 and 1 are not good because gcd(1, 2) = 1 and gcd(4, 1) = 1. As there are exactly 2 good elements, the permutation is beautiful.
The second sample has no beautiful permutations.
const int INF = 1000000000; const double eps = 1e-8; const int maxn = 300000; int ans[maxn]; int main() { //freopen("in.txt","r",stdin); int n,k; while(cin>>n>>k) { if(k > n - 1) { cout<<-1<<endl; continue; } repf(i,1,n) ans[i] = i; int Max = n - 1; int temp = Max - k; int t = temp/2; int e; for(int i = n;i>=2;i-=2) { if(t) { swap(ans[i],ans[i-1]); t--; }else { e = i; } } if(temp%2) { swap(ans[1],ans[e]); } printf("%d",ans[1]); repf(i,2,n) printf(" %d",ans[i]); cout<<endl; } return 0; }