題目

Given is an ordered deck of n cards numbered 1 to n with card 1 at the top and card n at the bottom. The following operation is performed as long as there are at least two cards in the deck:

Throw away the top card and move the card that is now on the top of the deck to the bottom of the deck.

Your task is to find the sequence of discarded cards and the last, remaining card.

Input

Each line of input (except the last) contains a number n ≤ 50. The last line contains ‘0’ and this line should not be processed.

Output

For each number from the input produce two lines of output. The first line presents the sequence of discarded cards, the second line reports the last remaining card. No line will have leading or trailing spaces. See the sample for the expected format.

Sample Input

7
19
10
6
0

Sample Output

Discarded cards: 1, 3, 5, 7, 4, 2
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 4, 8, 12, 16, 2, 10, 18, 14
Remaining card: 6
Discarded cards: 1, 3, 5, 7, 9, 2, 6, 10, 8
Remaining card: 4
Discarded cards: 1, 3, 5, 2, 6
Remaining card: 4

分析

這個(gè)題本身是個(gè)很簡單的模擬題，至少還剩兩張牌的時(shí)候?qū)⒌谝粡埲映鰜恚诙埛旁谧钕旅?，最后輸出把牌扔出來的序列和最后剩下的牌，我用stl庫里面的queue來模擬這個(gè)過程，本身很簡單的題，然而卻因?yàn)檩斎?的時(shí)候后面有空格而pe到死。

ac代碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;

 int main(){
    queue <int> qu;
    vector <int> ans; 
    int n;
        while(scanf("%d", &n) && n){
        for(int i = 1; i <= n; ++i){
            qu.push(i);
        }
        while(qu.size() > 1){
            ans.push_back(qu.front());
            qu.pop();
            qu.push(qu.front());
            qu.pop();
        }
        printf("Discarded cards:");
        bool text = 1;
        for(auto a : ans){
            if(text){
                text = 0;
            }
            else printf(",");
            printf(" %d", a);
        }
        printf("\n");
        printf("Remaining card: %d\n", qu.front());
        qu.pop();
        ans.clear();
    }
    return 0;
}