hdu5228 ZCC loves straight flush(模拟)

ZCC loves straight flush

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 611    Accepted Submission(s): 266

Problem Description

After losing all his chips when playing Texas Hold'em with Fsygd on the way to ZJOI2015, ZCC has just learned a black technology. Now ZCC is able to change all cards as he wants during the game. ZCC wants to get a Straight Flush by changing as few cards as possible. 

We call a five-card hand a Straight Flush when all five cards are consecutive and of the same suit. You are given a five-card hand. Please tell ZCC how many cards must be changed so as to get a Straight Flush.
  
Cards are represented by a letter('A', 'B', 'C', 'D') which denotes the suit and a number('1', '2', ⋯, '13') which denotes the rank.
  
Note that number '1' represents ace which is the largest actually. "1 2 3 4 5" and "10 11 12 13 1" are both considered to be consecutive while "11 12 13 1 2" is not.

 

Input

First line contains a single integer T(T=1000) which denotes the number of test cases.
For each test case, there are five short strings which denote the cards in a single line. It's guaranteed that all five cards are different.

 

Output

For each test case, output a single line which is the answer.

 

Sample Input

3A1 A2 A3 A4 A5A1 A2 A3 A4 C5A9 A10 C11 C12 C13

 

Sample Output

012

 

Source

BestCoder Round #41

 

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  解析：模拟。直接暴力枚举所有的胜利情况，然后一一匹配，得到匹配牌数x，需要改变的牌就为5-x。

             需要注意的是，5招牌并不一定是有序的，也就是说：A1 A2 A5 A4 A3 也是同花顺。一开始没注意到这个，调试一下午都没找出来。

代码：

#include<cstdio>#include<cstring>#include<algorithm>#define maxn 10using namespace std;int a[maxn],b[maxn];char s[maxn];void work(){  int i,j,k,p,q,sum,ans,x,y;  for(i=1;i<=5;i++)    {      scanf("%s",s);      a[i]=s[0]-'A';      b[i]=s[1]-'0';      k=strlen(s);      for(j=2;j<k;j++)b[i]=b[i]*10+s[j]-'0';    }    ans=5;  for(i=0;i<4;i++)    for(j=1;j<=10;j++)      {        sum=0,x=i,y=j;        for(k=0;k<5;k++)          {            if(y==14)y=1;            for(p=1;p<=5;p++)              if(a[p]==x && b[p]==y)sum++;            y++;            }        ans=min(ans,5-sum);        }  printf("%d\n",ans);  }int main(){  freopen("1.in","r",stdin);  int t,i,j,k;  while(scanf("%d",&t)!=EOF)    for(i=1;i<=t;i++)work();  return 0; }