Light OJ: 1235 – Coin Change (IV)

Solution Idea:

You generate all possible combinations for the left half and right half, and then sort one side, and for each element on the other side, search its corresponding one in the sorted side using binary search.

However, you have 3 choices for each item, i.e. take 0, 1, or 2 of it. So, if we try to calculate exactly, generating all possible combination for each side should take 3^(n/2) operations which is 3^9 in the worst case. let m = 3^(n/2), then the rest of the thing can be done in O(m lg m)

```
#include &lt;bits/stdc++.h&gt;

#define pii              pair &lt;int,int&gt;
#define pll              pair &lt;long long,long long&gt;
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout&lt;&lt;#x &quot; = &quot;&lt;&lt;(x)&lt;&lt;endl
#define VI               vector &lt;int&gt;
#define DBG              pf(&quot;Hi\n&quot;)
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf(&quot;%d&quot;,&amp;a)
#define sfl(a)           scanf(&quot;%lld&quot;,&amp;a)
#define sff(a,b)         scanf(&quot;%d %d&quot;,&amp;a,&amp;b)
#define sffl(a,b)        scanf(&quot;%lld %lld&quot;,&amp;a,&amp;b)
#define sfff(a,b,c)      scanf(&quot;%d %d %d&quot;,&amp;a,&amp;b,&amp;c)
#define sfffl(a,b,c)     scanf(&quot;%lld %lld %lld&quot;,&amp;a,&amp;b,&amp;c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i&lt;n;i++)
#define loop1(i,n)       for(int i=1;i&lt;=n;i++)
#define REP(i,a,b)       for(int i=a;i&lt;b;i++)
#define RREP(i,a,b)      for(int i=a;i&gt;=b;i--)
#define TEST_CASE(t)     for(int z=1;z&lt;=t;z++)
#define PRINT_CASE       printf(&quot;Case %d: &quot;,z)
#define CASE_PRINT       cout&lt;&lt;&quot;Case &quot;&lt;&lt;z&lt;&lt;&quot;: &quot;
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1&lt;&lt;28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1&lt;&lt;pos);}
//int reset(int N,int pos){return N= N &amp; ~(1&lt;&lt;pos);}
//bool check(int N,int pos){return (bool)(N &amp; (1&lt;&lt;pos));}
/*------------------------------------------------*/

ll coins[30];
ll n,k;

vector&lt;ll&gt;set1,set2;

void gen_set1(int idx, ll val)
{
set1.pb(val);
if(idx==n/2) return;
gen_set1(idx+1,val);
gen_set1(idx+1,val+coins[idx]);
gen_set1(idx+1,val+2*coins[idx]);
}

void gen_set2(int idx, ll val)
{
set2.pb(val);
if(idx==n) return;
gen_set2(idx+1,val);
gen_set2(idx+1,val+coins[idx]);
gen_set2(idx+1,val+2*coins[idx]);
}

int main()
{

///freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
///freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int t;
sf(t);
TEST_CASE(t)
{
sffl(n,k);
loop(i,n) sfl(coins[i]);
set1.clear();
set2.clear();

gen_set1(0,0);
gen_set2(n/2,0);

sort(all(set1));

PRINT_CASE;

bool test=0;

for(int i=0;i&lt;SZ(set2);i++)
{
if(binary_search(all(set1),k-set2[i]))
{pf(&quot;Yes\n&quot;);test=1;break;}
}

if(test==0)
pf(&quot;No\n&quot;);

}

return 0;
}

```