杭电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): 17556    Accepted Submission(s): 4439

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

 

 

Author

wangye

 

 

Source

HDU 2007-11 Programming Contest

 

 

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//二分法， 查找的是有序数组的下标，  

#include <iostream>
#include <algorithm>
using namespace std;
int a[550], b[550], c[550]; __int64 d[250050];
int main()
{
    int l,m ,n, cas=1;
    while(cin >> l >> m >> n)
    {
        
        int i, j, k, t=0;
        for(i=0; i<l; i++)
            cin >> a[i];
        for(i=0; i<m; i++)
            cin >> b[i];
        for(i=0; i<n; i++)
            cin >> c[i];
        for(i=0; i<l; i++)
            for(j=0 ;j<m; j++)
                d[t++] = a[i] + b[j];         
        sort(d, d+t);
        printf("Case %d:
", cas++);
        int g, sum;
        scanf("%d", &g);
        while(g--)
        {
            cin >> sum;
            int ok = 0;
            for(i=0; i<n; i++)
            {
                if(d[0] + c[i] <= sum && d[t-1] + c[i] >=sum)   //
                {
                    int l=0, r=t-1, mid;
                    while(r-l >= 0)
                    {
                        mid = (l+r)/2;
                        if(c[i] + d[mid] > sum) r = mid -1;
                        else if(d[mid] + c[i] < sum) l = mid + 1;
                        else
                        { ok = 1; break;} 
                    }
                    if(ok) break;
                }
            }
            if(ok)
            printf("YES
");
            else
            printf("NO
");
        }
    }
    return 0;
}

//感觉二分法做的确实有点懵。