背包dp（多重）

http://acm.hdu.edu.cn/showproblem.php?pid=1059

多重背包题；

如果sum奇数直接continue；不是奇数则判断dp[sum/2]能不能到达；

即dp[sum/2]的方案数是否为0；

注意输出格式！！！

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 const ll inf=0x3f3f3f3f;
 5 const int maxn=60000;
 6 long long dp[maxn],sum=0,val[maxn],cnt=0;
 7 signed main()
 8 {
 9     while(1)
10     {
11         int k=0;
12         sum=0;++cnt;
13         memset(val,0,sizeof(val));
14         memset(dp,0,sizeof(dp));
15         for(int i=1;i<=6;i++)
16         {
17             cin>>val[i];
18             sum+=val[i]*i;
19             if(val[i]) ++k;
20         }
21         if(!k) break;
22         cout<<"Collection #"<<cnt<<':'<<endl;
23         if(sum%2) {cout<<"Can't be divided."<<endl<<endl;continue;}
24         dp[0]=1;
25         for(ll i=1;i<=6;i++)
26         {
27             ll num=val[i];
28             for(ll h=1;num>0;h<<=1)
29             {
30                 if(h>num) h=num;
31                 num-=h;
32                 for(int j=sum/2;j>=i*h;j--)
33                     dp[j]+=dp[j-i*h];
34             }
35         }
36         if(dp[sum/2]) cout<<"Can be divided."<<endl;
37         else cout<<"Can't be divided."<<endl;
38         cout<<endl;
39     }
40 }

View Code