HDU ---- (5055) Bob and math problem (Greedy)

Bob and Math Problem

Time Limit: 2000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 401 accepted submission (s): 149

Problem descriptionrecently, Bob has been thinking about a math problem.
There are n digits, each digit is between 0 and 9. You need to use this n digits to constitute an integer.
This integer needs to satisfy the following conditions:

• 1. Must be an odd integer.

• 2. There is no leading zero.

• 3. Find the biggest one which is satisfied 1, 2.

Example:
There are three digits: 0, 1, 3. it can constitute six number of integers. only "301", "103" is legal, while "130", "310", "013", "031" is illegal. the biggest one of the odd integer is "301 ".

 

Inputthere are multiple test cases. Please process till EOF.
Each case starts with a line containing an integer N (1 <=n <= 100 ).
The second line contains N digits which indicate the digit $ A_1, A_2, A_3, \ cdots, a_n. (0 \ Leq a_ I \ Leq 9) $.

 

Outputthe output of each test case of a line. If you can constitute an integer which is satisfied above conditions, please output the biggest one. Otherwise, output "-1" instead.

 

Sample input30 1 335 4 232 4 6

 

Sample Output301425-1

 

Source bestcoder round #11 (Div. 2) greedy code:

 1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<functional> 5 #include<iostream> 6 using namespace std; 7 int str[120]; 8 int main() 9 {10   int n,i,res;11   while(scanf("%d",&n)!=EOF)12   {13       res=10;14       for(i=0;i<n;i++)15     {16        scanf("%d",&str[i]);17       if(str[i]&1==1&&res>str[i])18           res=str[i];19     }20     if(res==10){21         printf("-1\n");22         continue;23     }24     sort(str,str+n,greater<int>());25      int fir=-1,st=res;26     for(i=0;i<n;i++)27     {28         if(res==str[i]) res=-1;29         else30         {31             fir=str[i];32             break;33         }34     }35     if(fir==0) printf("-1\n");36     else37     {38         if(fir!=-1)39           printf("%d",fir);40         for( i++; i<n;i++)41         if(res==str[i])res=-1;42         else43          printf("%d",str[i]);44         printf("%d\n",st);45     }46   }47   return 0;48 }

View code

 

HDU ---- (5055) Bob and math problem (Greedy)