你好，母函数

#include <stdio.h>
#include <string.h>

int main( )
{
int m,n;
int i,j,k,t;
while(scanf("%d",&m)==1)
{
if(m<0) break;
n=0;
int val[m],num[m];//neng fang zhi shu zhu yue jian 
for( i=0;i<m;i++)
{
scanf("%d%d",&val[i],&num[i]);
n+=val[i]*num[i];
}
int c1[n+1],c2[n+1];
memset(c1,0,sizeof(c1));
memset(c2,0,sizeof(c2));
for(i=0;i<=num[0]*val[0]&&i<=n;i+=val[0])
{
c1[i]=1;
}
for(i=1;i<=m-1;i++)
{
for(j=0;j<=n;j++)
for(k=0;k<=num[i]*val[i]&&k+j<=n;k+=val[i])
c2[j+k]+=c1[j];
for(t=0;t<=n;t++)
{
c1[t]=c2[t];
c2[t]=0;
}
}
for(i=n/2;i>0;i--)
{
if(c1[i]!=0)
break;
}
printf("%d %d\n",n-i,i);
}
return 0;
}

下面是无限个物品的状态

#include <iostream>
using namespace std;
const int _max = 10001;
// c1是保存各项质量砝码可以组合的数目
// c2是中间量，保存每一次的情况

int c1[_max], c2[_max];
int main()
{    //int n,i,j,k;
int nNum;   //
int i, j, k;

while(cin >> nNum)
{
for(i=0; i<=nNum; ++i)
{
c1[i] = 1;
c2[i] = 0;
}
for(i=2; i<=nNum; ++i)
{

for(j=0; j<=nNum; ++j)
for(k=0; k+j<=nNum; k+=i)
{
c2[j+k] += c1[j];
}
for(j=0; j<=nNum; ++j)
{
c1[j] = c2[j];
c2[j] = 0;
}
}
cout << c1[nNum] << endl;
}
return 0;
}