Big Event in HDU (母函数， 玄学AC)

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).

InputInput 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.

OutputFor 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

母函数模板题

#include<bits/stdc++.h>

using namespace std;
#define maxn 1250000

int ans[maxn], temp[maxn];
int v[],num[];
int n, sum;

void init()
{
   // memset(a,0,sizeof(a));
   int mid = sum/;
    ans[]=;
    int i, j, k;
    for(i=; i<=num[]; i++)
        ans[i*v[]] = ;
    for(i=; i<n; i++)
    {
        for(j=; j<=mid; j++)
            for(k=; (k*v[i]+j)<=mid&&k<=num[i]; k++)
            {
                temp[j+k*v[i]] +=ans[j];
            }
        for(j=; j<=mid; j++)
        {
            ans[j] = temp[j];
            temp[j] = ;
        }
    }
}

int main()
{
    int n;
    while(cin >> n &&  n!= -)
    {
        sum = ;
        memset(ans, , sizeof(ans));
        for(int i = ; i < n; i++)
        {
            cin >> v[i] >> num[i];
            sum += v[i]*num[i];
        }
        init();
        for(int i = sum/; i>=; i--)
        {
            if(ans[i]) {cout << sum-i << " " << i << endl;break;}
        }
    }
    return ;
}

上面的代码连样例都过不了，但我们只要把 上面的 void init（） 函数写到主函数里面去， 就能AC， 楼主不太懂， 大佬能说一下吗？？？

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#define MAXN 1001000
using namespace std;

int value[200],number[200],ans[MAXN],temp[MAXN];
int main()
{
    int n,i,j,k;
    while(cin >> n&&n>=0)
    {
        int max = 0;
        for(i=0; i<n; i++)
        {
            cin >> value[i] >> number[i];
            max += value[i] * number[i];
        }
        int mid = max/2 ;
        memset(ans,0,sizeof(int)*mid+10);
        memset(temp,0,sizeof(int)*mid+10);
        for(i=0; i<=number[0]; i++)
            ans[i*value[0]] = 1;
        for(i=1; i<n; i++)
        {
            for(j=0; j<=mid; j++)
                for(k=0; (k*value[i]+j)<=mid&&k<=number[i]; k++)
                {
                    temp[j+k*value[i]] +=ans[j];
                }
            for(j=0; j<=mid; j++)
            {
                ans[j] = temp[j];
                temp[j] = 0;
            }
        }
        for(i=mid; i>=0; i--)
            if(ans[i]!=0)
                break;
        cout << max - i<<" "<< i << endl;
    }
    return 0;
}