zstu 1032 拆分物品 再01背包

判断能不能将总的物品分两半，中间拆分的时候有点多重背包的思想

#include<iostream>#include<string>#include<algorithm>using namespace std;int value[1005];int dp[60005];int num[7];int main(){    int i,j,d,ans,sum,ncase=1;    while(scanf("%d%d%d%d%d%d",&num[1],&num[2],&num[3],&num[4],&num[5],&num[6])!=EOF)    {         if(num[1]==0&&num[2]==0&&num[3]==0&&num[4]==0&&num[5]==0&&num[6]==0)            break;         d=sum=0;         for(i=1;i<=6;i++)         {             sum+=num[i]*i;             j=1;             while(j<=num[i])                                                                       {                  num[i]-=j;                  value[d++]=j*i;                  j*=2;             }             if(num[i]>0)                  value[d++]=num[i]*i;         }         printf("Collection #%d:\n",ncase++);         if(sum%2==1)         {             printf("Can't be divided.\n\n");             continue;         }         ans=sum/2;         memset(dp,0,sizeof(dp));         dp[0]=1;         for(i=0;i<d;i++)                                                                              for(j=ans;j>=value[i];j--)             if(dp[j-value[i]])                dp[j]=1;         if(dp[ans])            printf("Can be divided.\n\n");         else            printf("Can't be divided.\n\n");    }    return 0;}