文章目錄

• • 1.大賽題目
  • 2.中文翻譯
  • 3.代碼案例
  • 4.解題思路
  • • 4.1代碼舉例

1.大賽題目

Shuffle Card
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 6695 Accepted Submission(s): 2061

Problem Description
A deck of card consists of n cards. Each card is different, numbered from 1 to n. At first, the cards were ordered from 1 to n. We complete the shuffle process in the following way, In each operation, we will draw a card and put it in the position of the first card, and repeat this operation for m times.

Please output the order of cards after m operations.

Input
The first line of input contains two positive integers n and m.（1<=n,m<=105）

The second line of the input file has n Numbers, a sequence of 1 through n.

Next there are m rows, each of which has a positive integer si, representing the card number extracted by the i-th operation.

Output
Please output the order of cards after m operations. (There should be one space after each number.)

Sample Input
5 3
1 2 3 4 5
3
4
3

Sample Output
3 4 1 2 5

2.中文翻譯

洗牌。
時間限制：2000/1000毫秒（Java /其他）內存限制：65536/65536 K（Java /其他）
總提交量：6695份接受的提交量：2061份

問題

一副牌由N張牌組成。每一張牌都是不同的，編號從1到N。首先，牌是從1到N排列的。我們按照以下方式完成洗牌過程，在每一個操作中，我們將畫一張牌并將其放在第一張牌的位置上，然后重復操作m次。

請在M操作后輸出卡的順序。

輸入

輸入的第一行包含兩個正整數n和m。

輸入文件的第二行有n個數字，從1到n的序列。

接下來有m行，每行都有一個正整數si，表示第i個操作提取的卡號。

輸出

請在M操作后輸出卡的順序。（每個數字后面應該有一個空格。）

樣本輸入

5 3
1 2 3 4 5
3
4
3

樣本輸出

3 4 1 2 5

3.代碼案例

public class sixtest {public static void main(String args[]) {Scanner scanner = new Scanner(System.in);//n=5int n = scanner.nextInt();//m=3int m = scanner.nextInt();int A[] = new int[n];int out[] = new int[n];int B[] = new int[m];//輸入Afor(int i = 0;i<A.length;i++) {A[i] = scanner.nextInt();}//輸入Bfor(int i = 0;i<B.length;i++) {B[i] = scanner.nextInt();}//數組創造空間之后，默認每一個下標的值為0int Ex[] = new int [100010];int count = 0;//把要抽選的牌按照從后到前的順序裝進數組out中，重復的略過，然后再把剩余的裝進之后的數組下標for(int i = B.length-1; i >= 0; i--) {if(Ex[B[i]]==0) {out[count++] = B[i];//這里Ex[B[i]]等于幾都可以，只要不等于0Ex[B[i]] = 1;}}for(int i = 0;i<A.length;i++) {if(Ex[A[i]]==0) {out[count++] = A[i];}}for(int i = 0;i<out.length;i++) {System.out.print(out[i]+" ");}scanner.close();} }

4.解題思路

像下面這樣把要抽取的牌一張一張的放到最前面，后面的依次往后，可以抽相同的牌
->12345
->32145
->43125
->34125
把要抽選的牌按照從后到前的順序裝進數組out中，重復的略過，然后再把剩余的裝進剩余的數組下標

4.1代碼舉例

for(int i = B.length-1; i >= 0; i--) {if(Ex[B[i]]==0) {out[count++] = B[i];//這里Ex[B[i]]等于幾都可以，只要不等于0Ex[B[i]] = 1;} }

Ex[3] = 0->out[1] = 3->Ex[3] = 1
Ex[4] = 0->out[2] = 4->Ex[4] = 1
Ex[3] = 1 ->結束循環

for(int i = 0;i<A.length;i++) {if(Ex[A[i]]==0) {out[count++] = A[i];} }

Ex[1] = 0->out[3] = 1
Ex[2] = 0->out[4] = 2
Ex[3] = 1
Ex[4] = 1
Ex[5] = 0->out[5] = 5

總結

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