HDU 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): 3758    Accepted Submission(s): 935

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

 

 

这题是题好二分，要把两个数组合拼成一个。一个小地方错（代码里注明），让我找半天都找不到那里错。晕呀。。。现在还要搞清楚while(l < (=)r)的等号有与没的区别还有l = mid + 1和r = mid (- 1)要不要1。

还有如果数组从1...n和0...n-1的区别。要把这个弄清楚。

 

 

#include <stdio.h>
#include <stdlib.h>
#define maxn 505

int a[maxn], b[maxn], c[maxn], ab[maxn*maxn];
int l, n, m;

int Cmp(const void *a, const void *b)
{
    return *(int *)a - *(int *)b;
}

int F(int n, int k)
{
    int l, r, mid;
    l = 0;
    r = k - 1;//当时就是这里没多加个变量k,而直接l * m，搞得一直WA
    while (l < r)
    {
        mid = (l + r) / 2;
        if (ab[mid] == n)
        {
            return 1;
        }
        if (ab[mid] > n)
        {
            r = mid;
        }
        else
        {
            l = mid + 1;
        }
    }
    return 0;
}


int main()
{
    int   i, j, k, times = 1, flag, s, x;
    
    while (scanf("%d%d%d", &l, &m, &n) != EOF)
    {
        for (i = 0; i < l;scanf("%d", &a[i++]));
        for (j = 0; j < m;scanf("%d", &b[j++]));
        for (k = 0; k < n;scanf("%d", &c[k++]));
        
        
        k = 0;
        for (i = 0; i < l; i++)
        {
            for (j = 0; j < m; j++)
            {
                ab[k++] = a[i] + b[j];
            }
        }
        qsort(ab, k, sizeof(ab[0]), Cmp);

        printf("Case %d:\n", times++);
        scanf("%d", &s);

        while (s--)
        {
            scanf("%d", &x);
            flag = 1;
            for (i = 0; i < n; i++)
            {
                if (F(x - c[i], k) == 1)
                {
                    flag = 0;
                    break;
                }
            }
            if (flag == 0)
            {
                puts("YES");
            }
            else
            {
                puts("NO");
            }
        }
    }
    
    return 0;
}