HDU 4350 Card(数学)

Card

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 894    Accepted Submission(s): 518

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

113 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 34902908328 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

 

Source

2012 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int rec[100];int main(){int t,i,j,k=0;long long tmp;scanf("%d",&t);long long  a,b,c;while(t--){k++;for(i=1;i<=52;i++)scanf("%d",&rec[i]);cin>>a>>b>>c;tmp=a*(c-b+1);tmp%=c;for(i=1;i<=c;i++)if((i+tmp)%c==1)break;printf("Case #%d:",k);for(j=i;j<=c;j++)printf(" %d",rec[j]);for(j=1;j<i;j++)printf(" %d",rec[j]);for(i=c+1;i<=52;i++)printf(" %d",rec[i]);printf("\n");}return 0;}

这个题目比较简单了，就是一个简单的循环，根本就不需要什么模拟来实现，只要找到C前面的第一个应该输出的编号，输出到C，然后从第一个开始输出

最后输出C后面部分的就OK 了，题目比较简单！