hdu-1171-Big Event in HDU（背包）

Big Event in HDU

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 27   Accepted Submission(s) : 10

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 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0

 

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

 

题意：V是价值，M是数量 ，要求两个院系分的 价值差距最小。注意：结束输入的条件是小于等于0；

思路：相同价值的多个物品拆分成单个单个的物品 ，这样就 转化成01背包了 ，用价值限定 求出dp[sum/2] 的价值  即可。

code：

#include
#include
#include
using namespace std;
const int maxn = 101000;
int dp[maxn],v[maxn],m;
int main()
{
    //int n;
    while(scanf("%d",&m),m>0){
        memset(dp,0,sizeof(dp));
        memset(v,0,sizeof(v));
        int n=0,x,y,sum=0;
        for(int i=0;i=v[i];j--){
                dp[j]=max(dp[j],dp[j-v[i]]+v[i]);
            }
        }
        printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);

    }
    return 0;
}