040：Greedy Tino

Messj

Problem Description

      Tino wrote a long long story. BUT! in Chinese...
      So I have to tell you the problem directly and discard his long long story. That is tino want to carry some oranges with "Carrying pole", and he must make two side of the Carrying pole are the same weight. Each orange have its' weight. So greedy tino want to know the maximum weight he can carry.

Input

The first line of input contains a number t, which means there are t cases of the test data.
for each test case, the first line contain a number n, indicate the number of oranges.
the second line contains n numbers, Wi, indicate the weight of each orange.n is between 1 and 100, inclusive. Wi is between 0 and 2000, inclusive. the sum of Wi is equal or less than 2000.



Output
For each test case, output the maximum weight in one side of Carrying pole. If you can't carry any orange, output -1. Output format is shown in Sample Output.




Sample Input
1
5
1 2 3 4 5



Sample Output
Case 1: 7

Messj

1. #include<cstdio>
   
2. #define OFFSET 2000
   
3. #define INF 0x7fffffff
   
4. 
   
5. using namespace std;
   
6. int list[101];
   
7. int dp[101][4001];
   
8. 
   
9. int main()
   
10. {
    
11.         int T,n;
    
12.         int cas=0;
    
13.         bool have0;
    
14.         scanf("%d",&T);
    
15.         while(T--)
    
16.         {
    
17.                 cas++;
    
18.                 scanf("%d",&n);
    
19.                 have0=false;
    
20.                 int cnt=0;
    
21.                 for(int i=1;i<=n;i++)
    
22.                 {
    
23.                         scanf("%d",&list[++cnt]);
    
24.                         if(list[cnt]==0)
    
25.                         {
    
26.                                 cnt--;
    
27.                                 have0=true;
    
28.                         }
    
29.                 }
    
30.                 n=cnt;
    
31.                 for(int i=-2000;i<=2000;i++)
    
32.                         dp[0][i+OFFSET]=-INF;
    
33.                 dp[0][OFFSET]=0;
    
34.                 for(int i=1;i<=n;i++)
    
35.                 {
    
36.                         for(int j=-2000;j<=2000;j++)
    
37.                         {
    
38.                                 int a=-INF,b=-INF,c;
    
39.                                 if(j-list[i]>=-2000)
    
40.                                         a=dp[i-1][j-list[i]+OFFSET]+list[i];
    
41.                                 if(j+list[i]<=2000)
    
42.                                         b=dp[i-1][j+list[i]+OFFSET]+list[i];
    
43.                                 c=dp[i-1][j+OFFSET];
    
44.                                 a=a>b?a:b;
    
45.                                 dp[i][j+OFFSET]=a>c?a:c;
    
46.                         }
    
47.                        
    
48.                 }
    
49.                 if(dp[n][OFFSET]==0)
    
50.                         printf("Case %d: %d\n",cas,have0==true?0:-1);
    
51.                 else
    
52.                         printf("Case %d: %d\n",cas,dp[n][OFFSET]/2);
    
53.         }
    
54.         return 0;
    
55. }

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