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[501], b[501], c[501];
int sum[260000];
int L, N, M, S, X;
int cnt = 1;

int main()
{
    while(scanf("%d %d %d", &L, &N, &M) != EOF)
    {
        for(int i = 0; i < L; ++i)
            scanf("%d", &a[i]);
        for(int i = 0; i < N; ++i)
            scanf("%d", &b[i]);
        for(int i = 0; i < M; ++i)
            scanf("%d", &c[i]);
            
        int k = 0;
        for(int i = 0; i < L; ++i)
            for(int j = 0; 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 = 0;
            for(int i = 0; i < M; ++i)
            {
                int target = X - c[i];
                int left = 0, right = k;
                
                while(left <= right)
                {
                    int mid = (left + right) / 2;
                    if(sum[mid] == target)
                    {
                        flag = 1;
                        break;
                    }
                    else if(sum[mid] < target)
                        left = mid + 1;
                    else
                        right = mid - 1;
                }
                if(flag == 1)
                {
                    printf("YES\n");
                    break;
                }
            }
            if(flag == 0)
                printf("NO\n");
            
        }
    }
    

    
    return 0;
}