[置顶] Card

Card

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 664 Accepted Submission(s): 322

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 befor every number.

SampleInput

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

SampleOutput

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

AC CODE（15MS）

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
int T, ca = 1, N, n, l, r, a, i, cnt;
int card1[53], card2[53]; //分别记录card的当前状态和上一状态
scanf("%d", &T);
while(T--)
{
for(i = 1; i < 53; i++)
scanf("%d", &card1[i]);
scanf("%d%d%d", &n, &l, &r);
a = card1[1]; //记录首元素
N = n; //备份n
//cnt记录周期的长度
for(i = 1, cnt = 1; n--; i = -i, cnt++)
{
int j, k;
if(i > 0)
{
for(j = 1, k = l; k <= r; j++, k++)
card2[j] = card1[k];
for(k = 1; k < l; k++, j++)
card2[j] = card1[k];
for(k = r+1; k < 53; k++, j++)
card2[j] = card1[k];
if(a == card2[1])  n = N % cnt;  //找到周期
}
else
{
for(j = 1, k = l; k <= r; j++, k++)
card1[j] = card2[l+j-1];
for(k = 1; k < l; k++, j++)
card1[j] = card2[k];
for(k = r+1; k < 53; k++, j++)
card1[j] = card2[k];
if(a == card1[1])  n = N % cnt;  //找到周期
}
}
//输出
printf("Case #%d:", ca++);
if(i > 0)
{
for(int j = 1; j < 52; j++)
printf(" %d", card1[j]);
printf(" %d\n", card1[52]);
}
else
{
for(int j = 1; j < 52; j++)
printf(" %d", card2[j]);
printf(" %d\n", card2[52]);
}
}
return 0;
}