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🍀例题讲解2406:Card Stacking
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2406:Card Stacking
- 题目
- 思路
- AC代码
题目
总时间限制: 20000ms 单个测试点时间限制: 1000ms 内存限制: 65536kB
描述
Bessie is playing a card game with her N-1 (2 <= N <= 100) cow friends using a deck with K (N <= K <= 100,000; K is a multiple of N) cards. The deck contains M = K/N “good” cards and K-M “bad” cards. Bessie is the dealer and, naturally, wants to deal herself all of the “good” cards. She loves winning.
Her friends suspect that she will cheat, though, so they devise a dealing system in an attempt to prevent Bessie from cheating. They tell her to deal as follows:
-
Start by dealing the card on the top of the deck to the cow to her right
-
Every time she deals a card, she must place the next P (1 <= P <= 10) cards on the bottom of the deck; and
-
Continue dealing in this manner to each player sequentially in a counterclockwise manner
Bessie, desperate to win, asks you to help her figure out where she should put the “good” cards so that she gets all of them. Notationally, the top card is card #1, next card is #2, and so on.
输入
- Line 1: Three space-separated integers: N, K, and P
输出 - Lines 1…M: Positions from top in ascending order in which Bessie should place “good” cards, such that when dealt, Bessie will obtain all good cards.
样例输入
3 9 2
样例输出
3
7
8
提示
INPUT DETAILS:
Bessie is playing cards with two cow friends and a deck of 9 cards. She must place two cards on the bottom of the deck each time she deals one.
OUTPUT DETAILS:
Bessie should put the “good” cards in positions 3, 7, and 8. The cards will be dealt as follows; the card numbers are “position in original deck”:
Card Deck P1 P2 Bessie
Initial configuration 1 2 3 4 5 6 7 8 9 - - - - - - - - -
Deal top card [1] to Player 1 2 3 4 5 6 7 8 9 1 - - - - - - - -
Top card to bottom (#1 of 2) 3 4 5 6 7 8 9 2 1 - - - - - - - -
Top card to bottom (#2 of 2) 4 5 6 7 8 9 2 3 1 - - - - - - - -
Deal top card [4] to Player 2 5 6 7 8 9 2 3 1 - - 4 - - - - -
Top card to bottom (#1 of 2) 6 7 8 9 2 3 5 1 - - 4 - - - - -
Top card to bottom (#2 of 2) 7 8 9 2 3 5 6 1 - - 4 - - - - -
Deal top card [7] to Bessie 8 9 2 3 5 6 1 - - 4 - - 7 - -
Top card to bottom (#1 of 2) 9 2 3 5 6 8 1 - - 4 - - 7 - -
Top card to bottom (#2 of 2) 2 3 5 6 8 9 1 - - 4 - - 7 - -
Deal top card [2] to Player 1 3 5 6 8 9 1 2 - 4 - - 7 - -
Top card to bottom (#1 of 2) 5 6 8 9 3 1 2 - 4 - - 7 - -
Top card to bottom (#2 of 2) 6 8 9 3 5 1 2 - 4 - - 7 - -
Deal top card [6] to Player 2 8 9 3 5 1 2 - 4 6 - 7 - -
Top card to bottom (#1 of 2) 9 3 5 8 1 2 - 4 6 - 7 - -
Top card to bottom (#2 of 2) 3 5 8 9 1 2 - 4 6 - 7 - -
Deal top card [3] to Bessie 5 8 9 1 2 - 4 6 - 7 3 -
Top card to bottom (#1 of 2) 8 9 5 1 2 - 4 6 - 7 3 -
Top card to bottom (#2 of 2) 9 5 8 1 2 - 4 6 - 7 3 -
Deal top card [9] to Player 1 5 8 1 2 9 4 6 - 7 3 -
Top card to bottom (#1 of 2) 8 5 1 2 9 4 6 - 7 3 -
Top card to bottom (#2 of 2) 5 8 1 2 9 4 6 - 7 3 -
Deal top card [5] to Player 2 8 1 2 9 4 6 5 7 3 -
Top card to bottom (#1 of 2) 8 1 2 9 4 6 5 7 3 -
Top card to bottom (#1 of 2) 8 1 2 9 4 6 5 7 3 -
Deal top card [8] to Bessie - 1 2 9 4 6 5 7 3 8
Bessie will end up with the “good cards” that have been placed in positions 3, 7, and 8 in the original deck.
思路
看不懂题目的百度翻译一下谢谢
这个题目的要求就是n个人,k个牌,每次发牌完毕将顶部的p个牌放到末尾
我们直接循环
如果刚好等于本人,计入当前牌号
不是本人就出队
以上两个操作随便一个以后,循环出队入队完成放牌操作,具体看代码,一看就会!!
AC代码
#include<cmath>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int n,k,p;
queue<int>q;
int bs[100005];
int ecnt=0;
int main(){
cin>>n>>k>>p;
for(int i=1;i<=k;i++){
q.push(i);
}
for(int i=1;i<=k;i++){
if(i%n==0){
bs[++ecnt]=q.front();
}
q.pop();
for(int j=1;j<=p;j++){
int m=q.front();
q.push(m);
q.pop();
}
}
sort(bs+1,bs+1+ecnt);
for(int i=1;i<=ecnt;i++){
cout<<bs[i]<<endl;
}
}
