ACdreamOJ 1726 hash

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

Problem Description

Recently, Losanto find an interesting Math game. The rule is simple: Tell you a number H, and you can choose some numbers from a set {a[1],a[2],......,a[n]}.If the sum of the number you choose is H, then you win.
Losanto just want to know whether he can win the game.

Input

There are several cases.

In each case, there are two numbers in the first line n (the size of the set) and H. The second line has n numbers {a[1],a[2],......,a[n]}.0<n<=40, 0<=H<10^9, 0<=a[i]<10^9,All
the numbers are integers.

Output

If Losanto could win the game, output "Yes" in a line. Else output "No" in a line.

Sample Input

10 87
2 3 4 5 7 9 10 11 12 13
10 38
2 3 4 5 7 9 10 11 12 13

Sample Output

No
Yes

/**
ACdreamOJ  1726 hash
题目大意：给定n个数，问这些数的部分和是否可以组成m
解题思路：由于n的最大值只有40，因此我们把所有的数分成均等的两部分，利用哈希查寻一下就可以了
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <queue>
#include <map>
using namespace std;
typedef long long LL;
const int MAXN=1<<20;
const int HASH = 1000007;

struct hashmap//建立哈希表
{
    LL a[MAXN];
    int head[HASH],next[MAXN],size;
    void init() //初始化
    {
        memset(head,-1,sizeof(head));
        size=0;
    }
    bool find(LL val) //查找一个元素是否在哈希表内
    {
        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) //添加元素到哈希表中
    {
        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();
        for(int i=0; i<n; i++)
        {
            scanf("%d",&num[i]);
        }
        int t=n/2;
        for(int i=0; i<(1<<t); i++)
        {
            LL sum=0;
            for(int j=0; j<t; j++)
            {
                if(i&(1<<j))
                {
                    sum+=num[j];
                }
            }
            if(sum>m)continue;
            h1.add(sum);
        }
        int tt=n-t;
        int flag=0;
        for(int i=0; i<(1<<tt); i++)
        {
            LL sum=0;
            for(int j=0; j<tt; j++)
            {
                if(i&(1<<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;
}