hdu 4350 Card 规律题 多校联合赛事

Card

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 471Accepted Submission(s): 294

Problem Description

Bearchild is playing a card game with himself. But first of all, he needs to shuffle the cards. His strategy is very simple: After putting all the cards into a single stack,
he takes out some cards in the middle, from the L-th to the R-th when counting from top to bottom, inclusive, and puts them on the top. He repeats this action again
and again for N times, and then he regards his cards as shuffled.

Given L,R and N, and the initial card stack, can you tell us what will the card stack be like after getting shuffled?

Input

First line contains an integer T(1 <= T <= 1000), which is the test cases.
For each test case, first line contains 52 numbers(all numbers are distinct and between 1 and 52), which is the card number of the stack, from top to bottom.
Then comes three numbers, they are N, L and R as described. (0<=N<=10⁹, 1<=L<=R<=52)

Output

For each test case, output "Case #X:", X is the test number, followed by 52 numbers, which is the card number from the top to bottom.Note that you should output one and only
one blank before every number.

Sample Input

1
13 2 10 50 1 28 37 32 30 46 19 47 33 41 24 52 27 42 49 18 9 48 23 35 31 8 7 12 6 5 3 22 43 36 51 40 26 4 44 
17 39 38 15 14 25 16 29 20 21 45 11 34
902908328 38 50

Sample Output

Case #1: 26 4 44 17 39 38 15 14 25 16 29 20 21 45 13 2 10 50 1 28 37 32 30 46 19 47 33 41 24 52 27 42 49 18 
9 48 23 35 31 8 7 12 6 5 3 22 43 36 51 40 11 34

Author

elfness@UESTC_Oblivion

题意：一共52个数 输入n left right 表示把从left 到right的数拿出来 拿出来放到数串的最前面 如此重复n边 问最后的串是什么 （n很大）



思路 由于n 很大 貌似应该有规律 这时候可以找一个短点的小串 比如1 2 3 4 5 6 7 8 对这个串进行操作 最后可以发现

把一个串中的 从left 到right 移动 每移动right次 就会还原 故可对n进行%right 后 暴力

#include<stdio.h>
int main()
{
	int cas,i,n,left,right,pos,k;
	int a[55],change[55];
	scanf("%d",&cas);
	for(k=1;k<=cas;k++)
	{
           for(i=1;i<=52;i++)
			   scanf("%d",&a[i]);
		   scanf("%d %d %d",&n,&left,&right);
		   n=n%right;
		   while(n--)
		   {
		         pos=0;
		         for(i=left;i<=right;i++)
				 {
                        change[++pos]=a[i];
				 }
				 for(i=1;i<left;i++)
					 change[++pos]=a[i];
				 for(i=right+1;i<=52;i++)
					 change[++pos]=a[i];
				 for(i=1;i<=52;i++)
					 a[i]=change[i];
		   }
          printf("Case #%d:",k);
		   for(i=1;i<=52;i++)
			   printf(" %d",a[i]);
		   printf("\n");
	}
	return 0;
	
}