Acdream 1726 A Math game (binary enumeration +hash)

Topic Links:

http://acdream.info/problem?pid=1726

Test instructions

Given the number of N, and a number, judge whether K can be composed of any number of them.

Analysis:

Since n Max is 40, the brute force enumerates all the case complexity for 2^40 sure tle, and then comes to think of the binary enumeration

Divided into two halves, first to deal with the combination of the first N/2 number of cases, the results hash, and then enumerate the second half

And then find it in the hash table. Time complexity of O (2^ (N/2));

The code is as follows:

#include <stdio.h> #include <iostream> #include <cstdio> #include <cstring> #include < algorithm> #define PB push_back#define MP make_pair#define REP (i,n) for (int i=0;i< (n), ++i) #define for (I,L,H) for ( int i= (l); i<= (h); ++i) #define DWN (i,h,l) for (int i= (h); i>= (l);-i) #define CLR (Vis) memset (vis,0,sizeof (VIS)) # Define MST (Vis,pos) memset (vis,pos,sizeof (VIS)) #define MAX3 (A,B,C) max (A,max (b,c)) #define MAX4 (A,B,C,D) max (max (b) , Max (c,d)) #define MIN3 (a,b,c) min (a,min (b,c)) #define MIN4 (a,b,c,d) min (min (A, b), Min (c,d)) #define PI ACOs (-1.0) # Define INF 1000000000#define linf 1000000000000000000ll#define eps 1e-8#define LL long longusing namespace Std;const int M    axn=1<<20;const int hash = 1000007;struct hashmap//build hash Table {LL A[MAXN];    int head[hash],next[maxn],size;        void init () {//Initialize MST (head,-1);    size=0;        } bool Find (LL val) {//Find whether an element is in the hash table int tmp = (Val%hash + hash)%hash;           for (int i = head[tmp];i!=-1;i=next[i]) if (Val==a[i]) return true;    return false;        } void Add (LL val) {//Add element to hash table in int tmp = (val%hash+hash)%hash;        if (Find (Val)) return;        A[size]=val;        NEXT[SIZE]=HEAD[TMP];    head[tmp]=size++;        }}h1;int N,m,num[55];int Main () {while (~scanf ("%d%d", &n,&m)) {h1.init ();        REP (i,n) scanf ("%d", num+i);        int T=N/2;            REP (i,1<<t) {LL sum=0;            REP (j,t) {if (i& (1&LT;&LT;J)) sum+=num[j];            } if (sum>m) continue;        H1.add (sum);        } int tt=n-t;        int flag=0;            REP (I,1&LT;&LT;TT) {LL sum=0;            REP (J,TT) {if (i& (1&LT;&LT;J)) sum+=num[t+j];            } if (sum>m) continue;                if (H1.find (m-sum)) {flag=1;            Break        }} if (flag) printf ("yes\n"); else printf ("no\n"); } return 0;}

Acdream 1726 A Math game (binary enumeration +hash)