hdu 1171 多重背包

题意：给出价值和数量，求能分开的最近的两个总价值，例如10,20*2,30，分开就是40，40

链接：点我

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<queue>
 #include<map>
 using namespace std;
 #define MOD 1000000007
 const int INF=0x3f3f3f3f;
 const double eps=1e-;
 typedef long long ll;
 #define cl(a) memset(a,0,sizeof(a))
 #define ts printf("*****\n");
 const int MAXN=;
 int n,m,tt;
 int a[MAXN],b[MAXN];
 int dp[MAXN];

 int nValue;
 //0-1背包，代价为cost,获得的价值为weight
 void ZeroOnePack(int cost,int weight)
 {
     for(int i=nValue;i>=cost;i--)
         dp[i]=max(dp[i],dp[i-cost]+weight);
 }
 //完全背包，代价为cost,获得的价值为weight
 void CompletePack(int cost,int weight)
 {
     for(int i=cost;i<=nValue;i++)
         dp[i]=max(dp[i],dp[i-cost]+weight);
 }
 //多重背包
 void MultiplePack(int cost,int weight,int amount)
 {
     if(cost*amount>=nValue)CompletePack(cost,weight);
     else
     {
         int k=;
         while(k<amount)
         {
             ZeroOnePack(k*cost,k*weight);
             amount-=k;
             k<<=;
         }
         ZeroOnePack(amount*cost,amount*weight);
     }
 }
 int main()
 {
     int i,j,k;
     #ifndef ONLINE_JUDGE
     freopen("1.in","r",stdin);
     #endif
     while(scanf("%d",&n)!=EOF)
     {
         if(n<) break;
         nValue=;
         for(i=;i<n;i++)
         {
             scanf("%d%d",&a[i],&b[i]);
             nValue+=a[i]*b[i];
         }
         cl(dp);
         for(i=;i<n;i++)
         {
             MultiplePack(a[i],a[i],b[i]);
         }
         int v=nValue/;
         while(dp[v]!=v) v--;
         printf("%d %d\n",nValue-v,v);
     }
 }