(Codeforces Round #136 (Div. 2))A. Little Elephant and Function(递归分析,简单)
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The Little Elephant enjoys recursive functions.
This time he enjoys the sorting function. Let a is a permutation of an integers from 1 ton, inclusive, and ai denotes thei-th element of the permutation. The Little Elephant's recursive functionf(x), that sorts the first x permutation's elements, works as follows:
- If x = 1, exit the function.
- Otherwise, call f(x - 1), and then makeswap(ax - 1, ax) (swap thex-th and (x - 1)-th elements ofa).
The Little Elephant's teacher believes that this function does not work correctly. But that-be do not get an F, the Little Elephant wants to show the performance of its function. Help him, find a permutation of numbers from 1 ton, such that after performing the Little Elephant's function (that is callf(n)), the permutation will be sorted in ascending order.
A single line contains integer n (1 ≤ n ≤ 1000) — the size of permutation.
In a single line print n distinct integers from 1 ton — the required permutation. Numbers in a line should be separated by spaces.
It is guaranteed that the answer exists.
1
1
2
2 1
模拟一下即可。假设满足题意的数列为a1, a2, a3, a4,...,an,由an到a1依次执行递归函数后,得到的数列为: a2, a3, ...,an,a1.也就是说只是把原数列的头元素掉到最后面。递归函数要得到一个递增序列,就是1,2,3,...,n,那么原数列就自然是n,1,2,...,n-1,这样把头元素调到最后面就是所要求的序列。
AC CODE
#include <iostream>#include <string>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#include <vector>#define LL long long#define MAXI 2147483647#define MAXL 9223372036854775807#define eps (1e-8)#define dg(i) cout << "*" << i << endl;using namespace std;int main(){ int n; while(scanf("%d", &n) != EOF) { printf("%d ", n); for(int i = 1; i < n - 1; i++) printf("%d ", i); if(n != 1) printf("%d\n", n - 1); } return 0;}
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