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  • AtCoder Regular Contest 077 C

    题目链接:http://arc077.contest.atcoder.jp/tasks/arc077_a

    Time limit : 2sec / Memory limit : 256MB

    Score : 300 points

    Problem Statement

    You are given an integer sequence of length na1,…,an. Let us consider performing the following n operations on an empty sequence b.

    The i-th operation is as follows:

    1. Append ai to the end of b.
    2. Reverse the order of the elements in b.

    Find the sequence b obtained after these n operations.

    Constraints

    • 1≤n≤2×105
    • 0≤ai≤109
    • n and ai are integers.

    Input

    Input is given from Standard Input in the following format:

    n
    a1 a2  an
    

    Output

    Print n integers in a line with spaces in between. The i-th integer should be bi.


    Sample Input 1

    Copy
    4
    1 2 3 4
    

    Sample Output 1

    Copy
    4 2 1 3
    
    • After step 1 of the first operation, b becomes: 1.
    • After step 2 of the first operation, b becomes: 1.
    • After step 1 of the second operation, b becomes: 1,2.
    • After step 2 of the second operation, b becomes: 2,1.
    • After step 1 of the third operation, b becomes: 2,1,3.
    • After step 2 of the third operation, b becomes: 3,1,2.
    • After step 1 of the fourth operation, b becomes: 3,1,2,4.
    • After step 2 of the fourth operation, b becomes: 4,2,1,3.

    Thus, the answer is 4 2 1 3.


    Sample Input 2

    Copy
    3
    1 2 3
    

    Sample Output 2

    Copy
    3 1 2
    

    As shown above in Sample Output 1, b becomes 3,1,2 after step 2 of the third operation. Thus, the answer is 3 1 2.


    Sample Input 3

    Copy
    1
    1000000000
    

    Sample Output 3

    Copy
    1000000000
    

    Sample Input 4

    Copy
    6
    0 6 7 6 7 0
    

    Sample Output 4

    Copy
    0 6 6 0 7 7
    

     题解:找规律

     1 #include <iostream>
     2 #include <algorithm>
     3 #include <cstring>
     4 #include <cstdio>
     5 #include <vector>
     6 #include <cstdlib>
     7 #include <iomanip>
     8 #include <cmath>
     9 #include <ctime>
    10 #include <map>
    11 #include <set>
    12 #include <queue>
    13 #include <stack>
    14 using namespace std;
    15 #define lowbit(x) (x&(-x))
    16 #define max(x,y) (x>y?x:y)
    17 #define min(x,y) (x<y?x:y)
    18 #define MAX 100000000000000000
    19 #define MOD 1000000007
    20 #define pi acos(-1.0)
    21 #define ei exp(1)
    22 #define PI 3.141592653589793238462
    23 #define INF 0x3f3f3f3f3f
    24 #define mem(a) (memset(a,0,sizeof(a)))
    25 typedef long long ll;
    26 ll gcd(ll a,ll b){
    27     return b?gcd(b,a%b):a;
    28 }
    29 const int N=200005;
    30 const int mod=1e9+7;
    31 int a[N];
    32 int main()
    33 {
    34     std::ios::sync_with_stdio(false);
    35     int n;
    36     while(cin>>n){
    37         for(int i=1;i<=n;++i){
    38             cin>>a[i];
    39         }
    40         if(n&1){
    41             cout<<a[n];
    42             for(int i=n-2;i>=1;i-=2)
    43                 cout<<" "<<a[i];
    44             for(int i=2;i<=n-1;i+=2)
    45                 cout<<" "<<a[i];
    46             cout<<endl;
    47         }else{
    48             cout<<a[n];
    49             for(int i=n-2;i>=2;i-=2)
    50                 cout<<" "<<a[i];
    51             for(int i=1;i<=n-1;i+=2)
    52                 cout<<" "<<a[i];
    53             cout<<endl;
    54         }
    55     }
    56     return 0;
    57 }
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  • 原文地址:https://www.cnblogs.com/wydxry/p/7290437.html
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