【JZOJ6378】小w与数字游戏(game)

description

---

analysis

• 对于(n)很大，一眼看出来肯定有两个相等的数减出来是(0)，答案肯定是(0)

• 其实只要(n>7)，由于斐波那契数列，肯定能有几个数的和减去一个数凑出来(0)

• (n)很小就跑暴力

---

code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 300005
#define ha 19260817
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll a[MAXN];
ll n,T,ans;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline ll min(ll x,ll y){return x<y?x:y;}
inline void dfs(ll x,ll f[])
{
	if (x==n-1){ans=min(ans,f[1]);return;}
	ll g[15];fo(i,1,n)g[i]=0;
	fo(i,1,n-x-1)fo(j,i+1,n-x)
	{
		ll tot=1;
		fo(k,1,n-x)if (k!=i && k!=j)g[++tot]=f[k];
		g[1]=f[i]+f[j],dfs(x+1,g);
		g[1]=abs(f[i]-f[j]),dfs(x+1,g);
		g[1]=f[i]*f[j],dfs(x+1,g);
	}
}
int main()
{
	freopen("game.in","r",stdin);
	freopen("game.out","w",stdout);
	T=read();
	while (T--)
	{
		n=read(),ans=ha;
		fo(i,1,n)a[i]=read();
		if (n>7){printf("0
");continue;}
		dfs(0,a),printf("%lld
",ans);
	}
	return 0;
}