hdu1171 二进制优化背包问题

Big Event in HDU

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：给定n个物品，个数为num，价值为value，将总价值分成两部分，其实尽可能的接近。
思路：背包问题。dp[j]表示价值为j是否存在。
转移方程dp[j]|=dp[j-v[i]];
坑：从sum/2 开始找，如果是奇数则先输出小再输出大。

代码：

#include
#include
#include
using namespace std;
const int N=55,M=105;
int n,m;
int v[N*10];
int dp[N*N*M];
int main()
{
    while(scanf("%d",&n)!=EOF){
            if(n<0)break;
            int sum=0;
            int num,cnt=0,value;
            memset(dp,0,sizeof(dp));
            for(int i=1;i<=n;i++){
                scanf("%d%d",&value,&num);
                sum+=value*num;
                int t=1;
                while(num>t){
                    v[++cnt]=t*value;
                    num-=t;
                    t<<=1;
                }
                v[++cnt]=num*value;
            }
            dp[0]=1;
            for(int i=1;i<=cnt;i++)
                for(int j=sum;j>=v[i];j--){
                   dp[j]|=dp[j-v[i]];
                }
            for(int i=sum/2;i>=0;i--)//从中间往小找
            {
                if(dp[i]){
                    printf("%d %d\n",sum-i,i);
                    break;
                }
            }
          /*  int i;
            i=sum%2?(sum/2+1):(sum/2);//从中间往大的找要这么处理。
            for(;i<=sum;i++){
                if(dp[i]==1){
                    printf("%d %d\n",i,sum-i);
                    break;
                }
            }
            */
    }
    return 0;
}