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Card Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 889????Accepted Submission(s): 517 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 Source 2012 Multi-University Training Contest 6 Recommend zhuyuanchen520

思路：通過初結果到最終結果的變換中，r右邊的牌順序不變。然后左邊牌則按照順序插，所以只需要找到切牌后的第一張牌就完事了。

? ? ? ? ? ?而第一張牌的轉移方程為 第一張牌+l%(r-1)為下一次的。

#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int mm=110; int pai[mm]; int pos; int xun[12345]; void getxun(int s,int r) { int len=s-1;while(s!=1){xun[++pos]=s;s=(s+len)%r;if(s==0)s=r;}xun[++pos]=1; } int main() {int cas;int num,l,r;while(~scanf("%d",&cas)){for(int ca=1;ca<=cas;++ca){printf("Case #%d:",ca);for(int i=1;i<=52;++i)scanf("%d",&pai[i]);scanf("%d%d%d",&num,&l,&r);if(l==1){for(int i=1;i<=52;++i)printf(" %d",pai[i]);printf("\n");continue;}pos=0;getxun(l,r);num%=pos;if(num==0)num=pos;int kai=xun[num];for(int i=kai;i<=r;++i)printf(" %d",pai[i]);for(int i=1;i<kai;++i)printf(" %d",pai[i]);for(int i=r+1;i<=52;++i)printf(" %d",pai[i]);printf("\n");}}return 0; }

轉載于:https://www.cnblogs.com/nealgavin/p/3205890.html

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