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Dividing

1 0 1 2 0 0
1 0 0 0 1 1
0 0 0 0 0 0

Collection #1:
Can‘t be divided.

Collection #2:
Can be divided.

给出价值分别为1 2 3 4 5 6 的石头的个数 问能不能平分 

一个容量为sum/2的背包

#include<cstdio>
#include<cstring>
int maxn(int a,int b)
{
    return a>b?a:b;
}
int cnt[10];
int w[22222];
int dp[22222];
int main()
{
    int sum;
    int cas=1;
    int xx;
    while(scanf("%d",&cnt[0])!=EOF)
    {
        sum=cnt[0];
        for(int i=1;i<6;i++)
        {
            scanf("%d",&cnt[i]);
            sum+=cnt[i]*(i+1);
        }
        if(!sum)
            break;
        xx=0;
        for(int i=0;i<6;i++)
        {
            for(int j=1;j<=cnt[i];j*=2)
            {
                 w[xx++]=j*(i+1);
                 cnt[i]-=j;
            }
            if(cnt[i])
            {
                w[xx++]=cnt[i]*(i+1);
                cnt[i]=0;
            }
        }
        printf("Collection #%d:\n",cas++);
        //printf("%d\n",sum);
        if(sum%2)
        {
            printf("Can't be divided.\n");
        }
        else
        {
            memset(dp,0,sizeof(dp));
            for(int i=0;i<xx;i++)
            {
                for(int j=sum/2;j>=w[i];j--)
                {
                    dp[j]=maxn(dp[j],dp[j-w[i]]+w[i]);
                }
            }
            //printf("%d\n",dp[sum/2]);
            if(dp[sum/2]==sum/2)
            {
                printf("Can be divided.\n");
            }
            else
            {
                printf("Can't be divided.\n");
            }
        }
        puts("");
    }
    return 0;
}