UVA - 10935 Throwing cards away I
2014-08-13 14:31
302 查看
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.
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
题意: 桌子上面有一叠牌,从上到下依次编号为 1 - n; 做如下的操作:把第一张牌扔掉,把新的第一张牌放到最后一直操作:
#include <iostream> #include <algorithm> #include <cstdio> #include <queue> using namespace std; const int maxn = 100; queue<int> q; int main() { int n; int ans[maxn]; while(scanf("%d", &n) != EOF && n) { for(int i = 1; i <= n; i++) q.push(i); int k = 0; while(!q.empty()) { ans[k++] = q.front(); q.pop(); int t = q.front(); q.pop(); q.push(t); } printf("Discarded cards:"); for(int i = 0; i < n-1; i++) { if(i) printf(","); printf(" %d", ans[i]); } printf("\nRemaining card: %d\n", ans[n-1]); } return 0; }
相关文章推荐
- UVa 10935 Throwing cards away I【队列模拟】
- UVa 10935 Throwing cards away I(习题5-3)
- UVA 10935 - Throwing cards away I
- UVa---------10935(Throwing cards away I)
- 【习题5-3 UVA-10935】Throwing cards away I
- 习题5-3 UVA 10935 Throwing cards away I 卡片游戏
- Uva10935 - Throwing cards away I
- uva 10935 Throwing cards away I
- 经典第五章 习题 5-3 UVA 10935 Throwing cards away I(队列的简单应用)
- UVA 10935 - Throwing cards away I
- UVa 10935 Throwing cards away I
- UVa 10935 Throwing cards away I【队列】
- UVa 10935 - Throwing cards away I(模拟)
- UVa 10935 - Throwing cards away I STL
- uva 10935 Throwing cards away I
- Uva 10935 Throwing cards away I
- 卡片游戏(Throwing cards away I,UVa 10935)
- UVA 10935 Throwing cards away I
- Uva 10935 Throwing cards away I
- UVa 10935 - Throwing cards away I