Can you find it? HDU-2141 （二分查找模版题）

Description

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.

Input

There are many cases. Every data case is described as followed: In the first line there are three integers L, N, M, in the second line there are L integers represent the sequence A, in the third line there are N integers represent the sequences B, in the forth line there are M integers represent the sequence C. In the fifth line there is an integer S represents there are S integers X to be calculated. 1<=L, N, M<=500, 1<=S<=1000. all the integers are 32-integers.

Output

For each case, firstly you have to print the case number as the form "Case d:", then for the S queries, you calculate if the formula can be satisfied or not. If satisfied, you print "YES", otherwise print "NO".

Sample Input

3 3 3
1 2 3
1 2 3
1 2 3
3
1
4
10

Sample Output

Case 1:
NO
YES
NO

思路：枚举三个数组，时间复杂度O(N^3)，肯定会超时。所以可以把前两个数组的和先枚举出来，存在数组sum[260000]中，

由于要找是否存在sum[i] + c[j] = X, 可以把问题转换为在sum数组中查找是否有X-c[j]这个数。用二分查找sum数组即可。时间复杂度为： S * c数组的大小 * log(sum数组的大小)

注意：不能在c数组中二分查找是否有X-sum[i]这个数，因为其时间复杂度为： S * sum数组的大小 * log(c数组的大小)，由于题目中sum数组大小最大为250000，c数组大小最大为500，S为1000，所以会超时

 #include <iostream>

 #include <queue>

 #include <cstring>

 #include <cstdio>

 #include <string>

 #include <algorithm>

 using namespace std;

 int a[], b[], c[];

 int sum[];

 int L, N, M, S, X;

 int cnt = ;

 int main()

 {

     while(scanf("%d %d %d", &L, &N, &M) != EOF)

     {

         for(int i = ; i < L; ++i)

             scanf("%d", &a[i]);

         for(int i = ; i < N; ++i)

             scanf("%d", &b[i]);

         for(int i = ; i < M; ++i)

             scanf("%d", &c[i]);

         int k = ;

         for(int i = ; i < L; ++i)

             for(int j = ; j < N; ++j)

                 sum[k++] = a[i] + b[j];

         sort(sum, sum+k);

         scanf("%d", &S);

         printf("Case %d:\n", cnt++);

         while(S--)

         {

             scanf("%d", &X);

             int flag = ;

             for(int i = ; i < M; ++i)

             {

                 int target = X - c[i];

                 int left = , right = k;

                 while(left <= right)

                 {

                     int mid = (left + right) / ;

                     if(sum[mid] == target)

                     {

                         flag = ;

                         break;

                     }

                     else if(sum[mid] < target)

                         left = mid + ;

                     else

                         right = mid - ;

                 }

                 if(flag == )

                 {

                     printf("YES\n");

                     break;

                 }

             }

             if(flag == )

                 printf("NO\n");

         }

     }

     return ;

 }