好久不看动态规划，竟然连最简单的背包问题都解决不了了，，，，这让人情何以堪！！！！！！！！先把0-1背包看了看，还有个地方不是太懂，下午有时间再把完全背包看看。具体这道题来说，其实是比较好想的，就是分两种情况，分别用0-1背包和完全背包就可以解决了。刚开始wa了几次，一直找不到错误，最后竟然发现输出顺序反了。。。。。。题目：

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9885    Accepted Submission(s): 3437


Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0  
Input Input contains multiple test cases. Each test case starts with a number N (0 A test case starting with a negative integer terminates input and this test case is not to be processed.
 
Output For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
 
Sample Input

2
10 1
20 1
3
10 1 
20 2
30 1
-1

 
Sample Output

20 10
40 40

 
ac代码：

#include 
#include 
#include 
using namespace std;
struct good{
  int value;
  int num;
}aa[55];
int sum,total,dp[150000];
int max(int a,int b){
  return a>b?a:b;
}
void full_bag(int x){
for(int v=aa[x].value;v<=total;++v)
   dp[v]=max(dp[v],dp[v-aa[x].value]+aa[x].value);
}
void one_zerobag(int x){
for(int i=1;i<=aa[x].num;++i){
for(int v=total;v>=aa[x].value;v--)
  dp[v]=max(dp[v],dp[v-aa[x].value]+aa[x].value);
}
}
int main(){
// freopen("4.txt","r",stdin);
  int n;
  while(scanf("%d",&n)&&n>0){
  sum=0;
  for(int i=1;i<=n;++i){
  scanf("%d%d",&aa[i].value,&aa[i].num);
  sum+=(aa[i].value*aa[i].num);
  }
  total=sum/2;
  memset(dp,0,sizeof(dp));
  for(int i=1;i<=n;++i){
  if(aa[i].num*aa[i].value>=total)
  full_bag(i);
  else
  one_zerobag(i);
  }
  printf("%d %d\n",sum-dp[total],dp[total]);
  }
  return 0;
}