Big Event in HDU

2
10 1
20 1
3
10 1 
20 2
30 1
-1

20 10
40 40

题意  把一堆东西尽量分为两份  第一份不小于第二份

把全部东西的总价值s除以2  让它装尽量多的东西作为第二份  剩下的就是第一份了 

题目有个小坑点  是以负数作为结束条件的  不是-1  还有不要開始把s/=2  后来第一份又用s*2-d[s]  由于s/2*2不一定等于s了 

<span style="font-family:Arial Black;">#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 55,  V = 255000;
int d[V], val[N], num[N];
int main()
{
    int n;
    while (scanf ("%d", &n),n>0)
    {
        int s = 0;
        memset(d,0,sizeof(d));
        for (int i = 1; i <= n; ++i)
        {
            scanf ("%d%d", &val[i], &num[i]);
            s += val[i] * num[i];
        }
        
        for (int i = 1; i <= n; ++i)
        {
            for (int k = 1; k<=num[i]; k *=2)
            {
                num[i] -= k;
                for (int j = s / 2; j >= k * val[i]; --j)
                    d[j] = max (d[j], d[j - k * val[i]] + k * val[i]);
            }
            if (num[i] != 0)
                for (int j = s / 2; j >= num[i] * val[i]; --j)
                    d[j] = max (d[j], d[j - num[i] * val[i]] + num[i] * val[i]);
        }
        printf ("%d %d\n", s - d[s/2], d[s/2]);
    }
    return 0;
}
</span>