HDU1171：Big Event in HDU(多重背包)

Problem Description
Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).

Input
Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output
For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output
20 10
40 40

之前用01背包做过，这次再用多重背包做一次

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

const int MAX=100000;
int dp[MAX];
int c[MAX],w[MAX];
int v;

void ZeroOnePack(int cost,int wei)//01
{
    int i;
    for(i = v; i>=cost; i--)
    {
        dp[i] = max(dp[i],dp[i-cost]+wei);
    }
}

void CompletePack(int cost,int wei)//完全
{
    int i;
    for(i = cost; i<=v; i++)
    {
        dp[i] = max(dp[i],dp[i-cost]+wei);
    }
}

void MultiplePack(int cost,int wei,int cnt)//多重
{
    if(v<=cnt*cost)//如果总容量比这个物品的容量要小，那么这个物品可以直到取完，相当于完全背包
    {
        CompletePack(cost,wei);
        return ;
    }
    else//否则就将多重背包转化为01背包
    {
        int k = 1;
        while(k<=cnt)
        {
            ZeroOnePack(k*cost,k*wei);
            cnt = cnt-k;
            k = 2*k;
        }
        ZeroOnePack(cnt*cost,cnt*wei);
    }
}

int main()
{
    int n;
    while(~scanf("%d",&n),n>0)
    {
        int i,sum;
        v = 0;
        for(i = 0; i<n; i++)
        {
            scanf("%d%d",&c[i],&w[i]);
            v+=c[i]*w[i];
        }
        sum = v;
        v = v/2;
        memset(dp,0,sizeof(dp));
        for(i = 0; i<n; i++)
        {
            MultiplePack(c[i],c[i],w[i]);
        }
        printf("%d %d\n",sum-dp[v],dp[v]);
    }
    return 0;
}

补充：软件开发 , C++ ,

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