二分查找 HDOJ 2141 Can you find it?

题目传送门

 /*
      题意：给出一个数，问是否有ai + bj + ck == x
      二分查找：首先计算sum[l] = a[i] + b[j]，对于q，枚举ck，查找是否有sum + ck == x
 */
 #include <cstdio>
 #include <algorithm>
 #include <cmath>
 using namespace std;
 typedef long long ll;
 const int MAXN = 5e2 + ;
 const int INF = 0x3f3f3f3f;
 ll a[MAXN], b[MAXN], c[MAXN];
 ll sum[MAXN*MAXN];
 int tot;
 bool my_binary_search(int l, int r, ll k)   {
     while (l < r)   {
         int mid = (l + r) >> ;
         if (sum[mid] == k)  return true;
         else if (sum[mid] > k)  r = mid;
         else    l = mid + ;
     }
     return false;
 }
 int main(void)  {       //HDOJ 2141 Can you find it?
     //freopen ("HDOJ_2141.in", "r", stdin);
     int l, n, m, s, cas = ;
     while (scanf ("%d%d%d", &l, &n, &m) == )   {
         for (int i=; i<=l; ++i)    scanf ("%I64d", &a[i]);
         for (int i=; i<=n; ++i)    scanf ("%I64d", &b[i]);
         for (int i=; i<=m; ++i)    scanf ("%I64d", &c[i]);
         scanf ("%d", &s);
         printf ("Case %d:\n", ++cas);
         sort (a+, a++l);  sort (b+, b++n);  sort (c+, c++m);
         tot = ;
         for (int i=; i<=l; ++i)    {
             for (int j=; j<=n; ++j)    {
                 sum[++tot] = a[i] + b[j];
             }
         }
         sort (sum+, sum++tot);
         ll mn = a[] + b[] + c[], mx = a[l] + b[n] + c[m];
         while (s--) {
             ll q;  scanf ("%I64d", &q);
             if (q < mn || q > mx)   {
                 puts ("NO");    continue;
             }
             bool flag = false;
             for (int i=; i<=m; ++i)    {
                 int p = lower_bound (sum+, sum++tot, q - c[i]) - sum;
                 if (p <  || p > tot)   continue;
                 if (sum[p] + c[i] == q) {
                     flag = true;    puts ("YES");   break;
                 }
                 //if (my_binary_search (1, tot, q - c[i]))    {
                     //flag = true;    puts ("YES");   break;
                 //}
             }
             if (!flag)  {
                 puts ("NO");
             }
         }
     }
     return ;
 }