做法有兩種，一種是從頂向下構造解答樹。

第二種，從下向上構造解答樹。

本題解析採用自頂向下，每次列舉左子樹用到的子集，則右子樹就是剩下的子集。科普子集

#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
const int maxn = 6;
struct Tree{
	double L, R;
	Tree():L(0), R(0){}
};

vector<Tree> tree[1<<maxn];
double w[maxn], sum[1<<maxn], r;
int n, vis[1<<maxn];
void dfs(int subset){
	if(vis[subset]) return;
	vis[subset] = true;
	int haveChildren = false;
	for(int left = subset & (subset - 1); left; left = (subset) & (left - 1)){
		haveChildren = true;
		int right = subset^left;
		double d1 = sum[right] / sum[subset];
		double d2 = sum[left] / sum[subset];
		dfs(left); dfs(right);
		for(int i = 0; i < tree[left].size(); ++i){
			for(int j = 0; j < tree[right].size(); ++j){
				Tree t;
				t.L = max(tree[left][i].L + d1, tree[right][j].L - d2);
				t.R = max(tree[left][i].R - d1, tree[right][j].R + d2);
				if(t.L + t.R < r) tree[subset].push_back(t);
			}
		}
	}
	if(!haveChildren) tree[subset].push_back(Tree());
}
int main(){
	int T;
	scanf("%d", &T);
	while(T--){
		scanf("%lf%d", &r, &n);
		for(int i = 0; i < n; ++i) scanf("%lf", &w[i]);
		for(int i = 0; i < (1 << n); ++i){
			sum[i] = 0;
			tree[i].clear();
			for(int j = 0; j < n; ++j){
				if(i & (1 << j)) sum[i] += w[j];
			}
		}
    int root = (1<<n)-1;
    memset(vis, 0, sizeof(vis));
    dfs(root);

    double ans = -1;
    for(int i = 0; i < tree[root].size(); i++)
      ans = max(ans, tree[root][i].L + tree[root][i].R);
    printf("%.10lf\n", ans);
  }
  return 0;
}