HDU 1171 Big Event in HDU 【母函数】

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24858    Accepted Submission(s): 8747

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

 

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 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

 /*
题解：求组合数问题，用母函数屡试不爽。 
*/ 

#include
#include
int c1[250002],c2[250002],a[55],b[55];
int main()
{
    int n,i,j,k,s;
    while(scanf("%d",&n)!=EOF)
    {
        if(n<0) break;
        for(i=1,s=0; i<=n; i++)
        {
            scanf("%d %d",&a[i],&b[i]);//a[i]为value，b[i]为number 
            s+=a[i]*b[i];
        }
   //     printf("--%d\n",s);
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2)); 
        for(i=0; i<=a[1]*b[1]; i+=a[1])
        {
            c1[i]=1;
        }
        for(i=2; i<=n; i++)
        {
            for(j=0; j<=s; j++)
            {
                for(k=0; k+j<=s&&k/a[i]<=b[i]; k+=a[i])
                {
                    c2[j+k]+=c1[j];
                }
            }
            for(j=0; j<=s; j++)
            {
                c1[j]=c2[j];
                c2[j]=0;
            }
        }
        for(i=s/2; i>=0; i--)
        {
            if(c1[i]!=0)
            {
                printf("%d %d\n",s-i,i); break;
            }
        }
    }
    return 0;
}