Description
Given a list of N integers with absolute values no larger than 10 15, find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.Input
The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than 10 15 in absolute value and separated by a single space. The input is terminated with N = 0Output
For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.Sample Input
1 10 3 20 100 -100 0Sample Output
10 1 0 2 #include<cstdio> #include<algorithm> #include<map> using namespace std; typedef long long ll; ll a1[22],a2[22],d[2000003]; map<ll,int>mp;//去重 struct node { int num; ll cost; } c[2000003]; bool cmp(const node& a,const node& b) { return a.cost<b.cost; } ll abs1(ll x) { if(x<0)x=-x; return x; } int main() { int n; while(~scanf("%d",&n),n) { int i,j,k1=n/2,k2=n-k1,y=0; for(i=0; i<k1; i++) scanf("%lld",&a1[i]); for(i=0; i<k2; i++) scanf("%lld",&a2[i]); for(int t=0; t<1<<k1; t++) { ll cost=0;int num=0; for(i=0; i<k1; i++) { if(t>>i&1) { cost+=a1[i]; num++; } } if(mp[cost])c[mp[cost]].num=min(c[mp[cost]].num,num); else { mp[cost]=y; c[y].cost=cost; c[y++].num=num; } } sort(c,c+y,cmp); y--; for(i=0; i<=y; i++) d[i]=c[i].cost; ll mi=1e17; int sum=0; for(int t=0; t<1<<k2; t++) { ll cost=0; int num=0; for(i=0; i<k2; i++) { if(t>>i&1) { cost+=a2[i]; num++; } } int id; ll ta; id=lower_bound(d+1,d+y+1,-cost)-d; if(id<=y)//防止越界 { ta=abs1(cost+d[id]); num=num+c[id].num; if(ta<mi&&num)mi=ta,sum=num; else if(ta==mi&&num)sum=min(sum,num); num-=c[id].num; } if(id-1>=0)//判断两个边界情况 { id--; ta=abs1(cost+d[id]); num+=c[id].num; if(ta<mi&&num)mi=ta,sum=num; else if(ta==mi&&num)sum=min(sum,num); } } printf("%lld %d\n",mi,sum); mp.clear(); } return 0; }