Codeforces Round #395 (Div. 2) A B题题解

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本文解析了CodeForces平台上的两道竞赛题：Taymyriscallingyou和Timofeyandcubes。介绍了题目的背景、输入输出要求，并提供了详细的解题思路及代码实现。

题目链接：http://codeforces.com/contest/764

只做出了2题，C题很悲伤的没有看懂。。。

Comrade Dujikov is busy choosing artists for Timofey's birthday and is recieving calls from Taymyr from Ilia-alpinist.

Ilia-alpinist calls every n minutes, i.e. in minutes n, 2n, 3n and so on. Artists come to the comrade every m minutes, i.e. in minutes m, 2m, 3m and so on. The day is z minutes long, i.e. the day consists of minutes 1, 2, ..., z. How many artists should be killed so that there are no artists in the room when Ilia calls? Consider that a call and a talk with an artist take exactly one minute.

Input

The only string contains three integers — n, m and z (1 ≤ n, m, z ≤ 104).

Output

Print single integer — the minimum number of artists that should be killed so that there are no artists in the room when Ilia calls.

Examples

input
1 1 10

output
10

input
1 2 5

output
2

input
2 3 9

output
1

Note

Taymyr is a place in the north of Russia.

In the first test the artists come each minute, as well as the calls, so we need to kill all of them.

In the second test we need to kill artists which come on the second and the fourth minutes.

In the third test — only the artist which comes on the sixth minute.

思路：

水题

#include<bits/stdc++.h>
using namespace std;

int main(){
    int n,m,z;
    cin>>n>>m>>z;
    int cnt=0;
    for(int i=1;i<=z;i++){
        if(i%n==0&&i%m==0)
            cnt++;
    }
    cout<<cnt;
    return 0;
}

Young Timofey has a birthday today! He got kit of n cubes as a birthday present from his parents. Every cube has a number ai, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n - i + 1)-th. He does this while i ≤ n - i + 1.

After performing the operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105) — the number of cubes.

The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109), where ai is the number written on the i-th cube after Dima has changed their order.

Output

Print n integers, separated by spaces — the numbers written on the cubes in their initial order.

It can be shown that the answer is unique.

Examples

input
7
4 3 7 6 9 1 2

output
2 3 9 6 7 1 4

input
8
6 1 4 2 5 6 9 2

output
2 1 6 2 5 4 9 6

Note

Consider the first sample.

At the begining row was [2, 3, 9, 6, 7, 1, 4].
After first operation row was [4, 1, 7, 6, 9, 3, 2].
After second operation row was [4, 3, 9, 6, 7, 1, 2].
After third operation row was [4, 3, 7, 6, 9, 1, 2].
At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].

思路：

感觉写麻烦了..

大体思路是判断反转次数，奇数次就翻转，偶数次不翻转

#include<bits/stdc++.h>
using namespace std;

int num[200005];

int main() {
    int n;
    scanf("%d",&n);
    for(int i=1; i<=n; i++) {
        scanf("%d",&num[i]);
    }
    if(n%2==0) {
        for(int i=1; i<=n/2; i++) {
            if(i%2==0)
                printf("%d ",num[i]);
            else
                printf("%d ",num[n-i+1]);

        }
        for(int i=n/2+1; i<=n; i++) {
            if(i%2!=0)
                printf("%d ",num[i]);
            else
                printf("%d ",num[n-i+1]);

        }
    } else if(n%2!=0) {
        for(int i=1; i<=n/2; i++) {
            if(i%2==0)
                printf("%d ",num[i]);
            else
                printf("%d ",num[n-i+1]);

        }
        for(int i=n/2+1; i<=n; i++) {
            if(i%2==0)
                printf("%d ",num[i]);
            else
                printf("%d ",num[n-i+1]);

        }
    }
    return 0;
}