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hdu 4350 Card(递推循环节，3级)

2013-07-20 10:16 316 查看

2013暑期多校联合训练——80+高校，300+队伍，10000元奖金，敬请期待~

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：通过初结果到最终结果的变换中，r右边的牌顺序不变。然后左边牌则按照顺序插，所以只需要找到切牌后的第一张牌就完事了。

而第一张牌的转移方程为第一张牌+l%(r-1)为下一次的。

#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;
}

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