HD5228 ZCC loves straight flush

ZCC loves straight flush

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 541    Accepted Submission(s): 233

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

 

就是对ABCD四个分别寻找，数据少枚举一下就可以了。第一次做这个题居然还纠结这么久

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        char card;        int a[15],b[15],c[15],d[15];        int n;        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(c,0,sizeof(c));        memset(d,0,sizeof(d));        for(int i=1;i<=5;i++)        {         // cin>>card>>n;          scanf("%c%d",&card,&n);          if(card=='A') a[i]=n;          if(card=='B') b[i]=n;          if(card=='C') c[i]=n;          if(card=='D') d[i]=n;          getchar();        }        //for(int i=1;i<10;i++)       //  cout<<a[i]<<" ";        int maxn=0;        for(int i=1;i<=5;i++)        {            int count1=0,count2=0;            if(a[i]>=1&&a[i]<10)            {                for(int j=1;j<=5;j++)                {                    if(a[j]-a[i]>=1&&a[j]-a[i]<=4&&a[j])                    count1++;                }            }            if(a[i]>=10)            {                for(int j=1;j<=5;j++)                {                    if((a[j]-a[i]>=1&&a[j]-a[i]<=4)||a[j]==1)                        count1++;                }            }            maxn=max(maxn,count1);        }        for(int i=1;i<=5;i++)        {            int count1=0;            if(b[i]>=1&&b[i]<10)            {                for(int j=1;j<=5;j++)                {                    if(b[j]-b[i]>=1&&b[j]-b[i]<=4)                        count1++;                }            }            if(b[i]>=10)            {                for(int j=1;j<=5;j++)                {                    if((b[j]-b[i]>=1&&b[j]-b[i]<=4)||b[j]==1)                        count1++;                }            }            maxn=max(maxn,count1);        }        for(int i=1;i<=5;i++)        {            int count1=0;            if(c[i]>=1&&c[i]<10)            {                for(int j=1;j<=5;j++)                {                    if(c[j]-c[i]>=1&&c[j]-c[i]<=4)                        count1++;                }            }            if(c[i]>=10)            {                for(int j=1;j<=5;j++)                {                    if((c[j]-c[i]>=1&&c[j]-c[i]<=4)||c[j]==1)                        count1++;                }            }            maxn=max(maxn,count1);        }        for(int i=1;i<=5;i++)        {            int count1=0;            if(d[i]>=1&&d[i]<10)            {                for(int j=1;j<=5;j++)                {                    if(d[j]-d[i]>=1&&d[j]-d[i]<=4)                        count1++;                }            }            if(d[i]>=10)            {                for(int j=1;j<=5;j++)                {                    if((d[j]-d[i]>=1&&d[j]-d[i]<=4)||d[j]==1)                        count1++;                }            }            maxn=max(maxn,count1);        }         printf("%d\n",4-maxn);    }    return 0;}