poj 1041Dividing解题报告可行性背包问题

题意是说，给一些大理石，每种有num[i]个，判断能否分成两部分，可行性问题，感觉不像是动态规划，中间没有最优子结构这个性质，倒像是纯粹的递推

#include<stdio.h>
#include<string.h>
#define N 60005
int bag[7];
int v[N];
bool dp[N];
int main()
{
    int sum;
    int i,j,k,ans;
    int icase=1;
    while(1)
    {
        sum=0;
        for(i=1;i<=6;i++)
        {
            scanf("%d",&bag[i]);
            sum+=bag[i]*i;
        }
        if(!sum)
        break;
        printf("Collection #%d:\n",icase++);
        if(sum&1)
        {
            printf("Can't be divided.\n\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        ans=sum>>1;
        k=0;
        for(i=1;i<=6;i++)
        {
            j=1;
            while(j<=bag[i])
            {
                bag[i]-=j;
                v[k++]=j*i;
                j<<=1;
            }
            if(bag[i])
            v[k++]=bag[i]*i;
        }
        dp[0]=true;
        for(i=0;i<k;i++)
        for(j=ans;j>=v[i];j--)
        if(dp[j-v[i]])
        dp[j]=true;
        if(dp[ans])
        printf("Can be divided.\n\n");
        else
        printf("Can't be divided.\n\n");
    }
    return 0;
}