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Language: Default Shuffle‘m Up
Description A common pastime for poker players at a poker table is to shuffle stacks of chips. Shuffling chips is performed by starting with two stacks of poker chips, S1 and S2, each stack containing C chips. The actual shuffle operation is performed by interleaving a chip from S1 with a chip from S2 as shown below for C = 5:
The single resultant stack, S12, contains 2 * C chips. The bottommost chip of S12 is the bottommost chip from S2. On top of that chip, is the bottommost After the shuffle operation, S12 is split into 2 new stacks by taking the bottommost C chips from S12 to form a new S1 and the topmost C chips For this problem, you will write a program to determine if a particular resultant stack S12 can be formed by shuffling two stacks some number of times. Input The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow. Each dataset consists of four lines of input. The first line of a dataset specifies an integer C, (1 ≤ C ≤ 100) which is the number of chips in each initial stack (S1 and S2). Output Output for each dataset consists of a single line that displays the dataset number (1 though N), a space, and an integer value which is the minimum number of shuffle operations required to get the desired resultant stack. If the Sample Input 2 4 AHAH HAHA HHAAAAHH 3 CDE CDE EEDDCC Sample Output 1 2 2 -1 Source |
题意：
已知两堆牌s1和s2的初始状态，其牌数均为c，依次交替叠放组合成一堆牌s12，再将s12的最底下的c块牌归为s1，最顶的c块牌归为s2，依此循环下去。
现在输入s1和s2的初始状态 以及最终状态s12
问s1 s2经过多少次洗牌之后，最终能达到状态s12，若永远不可能相同，则输出"-1"。直接模拟过程即可。
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 2005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int main()
{
int N,C,cas=1;
char s1[110],s2[110],str[222],s[222];
scanf("%d",&N);
while (N--)
{
map<string,int>q;
scanf("%d",&C);
scanf("%s%s%s",s1,s2,str);
bool ok=true;
int ans=0;
while (1)
{
int num=0,len1=0,len2=0;
while (num<2*C)
{
if (num%2)
s[num++]=s1[len1++];
else
s[num++]=s2[len2++];
}
s[num]='\0';
ans++;
if (strcmp(str,s)==0)
break;
if (q[s])
{
ok=false;
break;
}
q[s]=1;
for (int i=0;i<C;i++)
s1[i]=s[i];
s1[C]='\0';
for (int i=C;i<2*C;i++)
s2[i-C]=s[i];
s2[C]='\0';
}
if (ok)
printf("%d %d\n",cas++,ans);
else
printf("%d -1\n",cas++);
}
return 0;
}
/*
2
4
AHAH
HAHA
HHAAAAHH
3
CDE
CDE
EEDDCC
*/
Shuffle'm Up (poj 3087 模拟)