HDOJ 2141 Can you find it?

Can you find it?

Time Limit: 10000/3000 MS (Java/Others)    Memory Limit: 32768/10000 K (Java/Others)
Total Submission(s): 20106    Accepted Submission(s): 5099

Problem 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
   
   
   
   

 

用三个for循环会超时

后来就改进了一下把函数改为：A+B=X-C，然后二分搜一下就可以了；

#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 500 + 10;

long long ab[maxn * maxn], a[maxn], b[maxn], c[maxn];
long long len;
int binary_search(long long kk)
{
    long long fir = 0, las = len - 1;
    while (fir <= las){
        long long middle = (fir + las) / 2;
        if (ab[middle] == kk)
            return 1;
        else if (ab[middle] > kk)
            las = middle - 1;
        else
            fir = middle + 1;
    }
    return 0;
}

int main()
{
    long long l, n, m, s, x;
    long long ans, flag;
    ans = 0;
    while (scanf("%I64d%I64d%I64d", &l, &n, &m) != EOF){
        len = 0;
        for (int i = 0; i < l; i++)
            scanf("%I64d", &a[i]);
        for (int i = 0; i < n; i++)
            scanf("%I64d", &b[i]);
        for (int i = 0; i < m; i++)
            scanf("%I64d", &c[i]);

        for (int i = 0; i < l; i++){
            for (int j = 0; j < n; j++){
                ab[len++] = a[i] + b[j];
            }
        }
        sort (ab, ab + len);
        sort (c, c + m);
        scanf("%I64d", &s);
        printf("Case %I64d:\n", ++ans);
        while (s--){
            scanf("%I64d", &x);
            if (x < ab[0] + c[0] || x > ab[len - 1] + c[m - 1]){
                printf("NO\n");
                continue;
            }
            flag = 0;
            for (int i = 0; i < m; i++){
                long long kk = x - c[i];
                if (binary_search(kk)){
                    flag = 1;
                    break;
                }
            }
            if (flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}