HDU 1171 Big Event in HDU 多重背包

这题是我第一次写多重背包，1RE，1WA，

#include<stdio.h>
#include<string.h>
int n,v[55],num[55],dp[250000],sum,all;
int max( int a,int b )
{
    return a > b ? a : b;
}
int main( )
{
    while( scanf( "%d",&n ),n > 0 )
    {
           sum = 0;
           for( int i = 0; i < n; ++i )
                scanf( "%d%d",&v[i],&num[i] ),sum += v[i] * num[i];
           memset( dp,0,sizeof( dp ) );
           all = sum;
           sum /= 2;
           for( int i = 0; i < n; ++i )
                for( int j = 0 ; j < num[i]; ++j )
                     for( int k = sum; k >= v[i]; k -= v[i] )
                          dp[k] = max( dp[k-v[i]]+v[i],dp[k] );
           if( dp[sum] > all - dp[sum] )
               printf( "%d %d\n",dp[sum],all - dp[sum] );
           else
               printf( "%d %d\n",all - dp[sum],dp[sum] );
           }
    return 0;
}

下面这个是标准写法

#include<stdio.h>
#include<string.h>
int n,v[55],num[55],all,sum,dp[250005];
inline int max( int a,int b )
{
    return a > b ? a : b;
}
void zo( int v,int w )
{
     for( int j = sum; j >= v; --j )
          dp[j] = max( dp[j-v] + w,dp[j] );
 }
void co( int v,int w )
{
     for( int j = v; j <= sum; ++j )
          dp[j] = max( dp[j-v] + w,dp[j] );
 }
void mu( int v,int w,int c )
{
     if( v * c > sum )
         co( v,w );
     else
     {
         int k = 1;
         while( c > k )
         {
                zo( k * v,k * w );
                c -= k;
                k <<= 1;
                }
         zo( c * v,c * w );
     }
 }
int main( )
{
    while( scanf( "%d",&n ), n > 0 )
    {
           all = 0;
           memset( dp,0,sizeof( dp ) );
           for( int i = 0; i < n; ++i )
                scanf( "%d%d",&v[i],&num[i] ),all += v[i]*num[i];
           sum = all / 2;
           for( int i = 0; i < n; ++i )
                mu( v[i],v[i],num[i] );
           if( dp[sum] > all - dp[sum] )
               printf( "%d %d\n",dp[sum],all - dp[sum] );
           else
               printf( "%d %d\n",all - dp[sum],dp[sum] );
           }
    return 0;
}