题解：因为天平两边都可以放砝码，所以需要再建一个负的砝码值，然后用母函数即可……
#include <cstdio>
#include <iostream>
using namespace std;
int c1[10005],c2[10005],n,sum,res,weight[105];
int main(){
while(scanf("%d",&n)!=EOF){
res=sum=0;
for(int i=0;i<n;i++){
scanf("%d",&weight[i]);
sum+=weight[i];
weight[n+i]=-weight[i];
}
memset(c1,0,sizeof(c1)); memset(c2,0,sizeof(c2));
for(int i=0;i<=weight[0];i+=weight[0])c1[i]=1;
for(int i=1;i<2*n;i++){
for(int j=0;j<10005;j++){
if(i<=n-1){
for(int k=0;k<=weight[i];k+=weight[i])c2[j+k]+=c1[j];
}else{
for(int k=weight[i];k<=0;k+=-weight[i]){
if(j+k>=0)c2[j+k]+=c1[j];
}
}
}for(int j=0;j<10005;j++)c1[j]=c2[j],c2[j]=0;
}for(int i=0;i<=sum;i++)if(c1[i]==0)res++;
bool flag=false;
printf("%d
",res);
if(res!=0){
for(int i=0;i<=sum;i++){
if(c1[i]==0&&flag==false){
printf("%d",i);
flag=true;
}
else if(c1[i]==0&&flag)printf(" %d",i);
}printf("
");
}
}return 0;
}