Throwing cards away I
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
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Description
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.
Each line of input (except the last) contains a number n ≤ 50. The last line contains 0 and this line should not be processed. 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
Output for sample input
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
Folklore, adapted by Piotr Rudnicki
<span style="color:#6600cc;">/************************************************
author : Grant Yuan
time : 2014.8.4
algorithm : 约瑟夫环
source ; UVA 10935
************************************************/
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
queue<int> q;
int n;
int main()
{
while(1){
scanf("%d",&n);
if(n==0) break;
while(!q.empty()) q.pop();
for(int i=1;i<=n;i++)
{
q.push(i);
}
int t;
printf("Discarded cards:");
while(q.size()>1){
if(q.size()==n) printf(" %d",q.front());
else printf(", %d",q.front());
q.pop();
t=q.front();
q.pop();
q.push(t);
}
cout<<endl;
printf("Remaining card: %d\n",q.front());}
return 0;
}
</span>
UVA 10935 约瑟夫环