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 Description Recently, 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 odd Integer is "301".
Input There 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 $a_1, a_2, a_3, \cdots, a_n. ( 0 \leq a_i \leq 9)$.
Output The 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 Input 3 0 1 3 3 5 4 2 3 2 4 6
Sample Output 301 425 -1
Source BestCoder Round #11 (Div. 2) 贪心 代码：
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 else
30 {
31 fir=str[i];
32 break;
33 }
34 }
35 if(fir==0) printf("-1\n");
36 else
37 {
38 if(fir!=-1)
39 printf("%d",fir);
40 for( i++; i<n;i++)
41 if(res==str[i])res=-1;
42 else
43 printf("%d",str[i]);
44 printf("%d\n",st);
45 }
46 }
47 return 0;
48 }
View Code
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