HDU 2141 Can you find it?

题意：Give you three sequences of numbers A, B, C, then we give you a number X. Now you need to calculate if you can find the three numbers Ai, Bj, Ck, which satisfy the formula Ai+Bj+Ck = X.

即给出一个A数组，B数组，C数组，和S个询问，每个询问给出一个X值，问是否存在符合条件的等式。

分析：可以用hash存下a数组和b数组中所有元素的和，然后可以用hash或者二分的方法查找是否存在x-ck即可。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 505;
const int mod = 600000;
const int INF = (1<<30);
struct node
{
    int key;
    int count;
}h[mod];
int a[maxn];
int b[maxn];
int c[maxn];

void add(int p)
{
    int k = p%mod;
    if (k < 0)
        k += mod;
    while (h[k].count != 0)
    {
        if (h[k].key == p)
        {
            h[k].count++;
            break;
        }
        k = (k+1)%mod;
    }
    if (h[k].count == 0)
    {
        h[k].count++;
        h[k].key = p;
    }
}
bool find(int p)
{
    int k = p%mod;
     if (k < 0)
        k += mod;
    while (h[k].count!=0)
    {
        if (h[k].key == p)
            return true;
        k = (k+1)%mod;
    }
    return false;
}
int main()
{
    int i, j,k;
    int l, n, m, s;
    int p;
    int ca = 1;
    while (scanf("%d %d %d",&l,&m,&n)!=EOF)
    {
          memset(h,0,sizeof(h));
          for (i=0; i<l; i++)
            scanf("%d",&a[i]);
          for (i=0; i<m; i++)
            scanf("%d",&b[i]);
          for (i=0; i<n; i++)
            scanf("%d",&c[i]);
          sort(a,a+l);
          sort(b,b+m);
          sort(c,c+n);
          for (i=0; i<l; i++)
            for (j=0; j<m; j++)
                add(a[i]+b[j]);
          bool flag=false;
          scanf("%d",&s);
          printf("Case %d:\n",ca++);
          while (s--)
          {
             int p;
             scanf("%d",&p);
           /*  if (a[0]+b[0]+c[0]>p || a[l-1]+b[m-1]+c[n-1]<p)
            {
                printf("NO\n");
                continue;
            }*/
             for (i=0; i<n; i++)
                if (find(p-c[i]))
                    break;
             printf("%s\n",i==n?"NO":"YES");
          }
    }
    return 0;
}