Codeforces Round #175 (Div. 2) A. Slightly Decreasing Permutations(构造,简单)

A. Slightly Decreasing Permutations
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Permutation p is an ordered set of integers p1,  p2,  ...,  pn, consisting of n distinct positive integers, each of them doesn't exceed n. We'll denote the i-th element of permutation p as pi. We'll call number n the size or the length of permutation p1,  p2,  ...,  pn.

The decreasing coefficient of permutation p1, p2, ..., pn is the number of such i (1 ≤ i < n), that pi > pi + 1.

You have numbers n and k. Your task is to print the permutation of length n with decreasing coefficient k.

Input

The single line contains two space-separated integers: n, k (1 ≤ n ≤ 105, 0 ≤ k < n) — the permutation length and the decreasing coefficient.

Output

In a single line print n space-separated integers: p1, p2, ..., pn — the permutation of length n with decreasing coefficient k.

If there are several permutations that meet this condition, print any of them. It is guaranteed that the permutation with the sought parameters exists.

Sample test(s)
Input
5 2
Output
1 5 2 4 3
Input
3 0
Output
1 2 3
Input
3 2
Output
3 2 1
 1 #include <iostream>
 2 #include <fstream>
 3 #include <string>
 4 #include <set>
 5 #include <map>
 6 #include <vector>
 7 #include <stack>
 8 #include <queue>
 9 #include <cmath>
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13 #include <utility>
14 using namespace std;
15 #define ll long long
16 #define cti const int
17 #define ctll const long long
18 #define dg(i) cout << '*' << i << endl;
19 
20 int main()
21 {
22     int n, k;
23     while(scanf("%d %d", &n, &k) != EOF)
24     {
25         if(k == 0)
26         {
27             for(int i = 1; i < n; i++) cout << i << ' ';
28             cout << n << endl;
29             continue;
30         }
31         cout << n--;
32         while(k-- > 1) cout << ' ' << n--;
33         for(int i = 1; n--; i++) cout << ' ' << i;
34         cout << endl;
35     }
36     return 0;
37 }
原文地址:https://www.cnblogs.com/cszlg/p/2987915.html