HDU 2141 Can you find it?

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

 

题意：

给定三个数组a，b，c，每个数组有若干个数（<=500个），再给定一个数s要你判断是否存在s=a[i]+b[j]+c[k]，存在一组数就输出YES，一组都不存在就输出NO。

思路：

A+B = X - C

先把a数组和b数组中的数相加成一个ab[500×500]的数组，这样就相当于ab[i]+c[j]=s;再变形一下，ab[i]=c[j]+s，只要在ab数组里用二分查找看能否把c[j]+s找出来。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 505

__int64 ab[N * N];
int num;

bool _search(__int64 x)
{
    int i = 0, l = num - 1;
    int mid;
    while(i <= l){
        mid = (i + l) / 2;
        if(ab[mid] == x)
            return true;
        else if(ab[mid] < x)
            i = mid + 1;
        else
            l = mid - 1;
    }
    return false;
}

int main()
{
    int n, m, l, kase = 0, s;
    __int64 a[N], b[N], c[N], x;
    while(scanf("%d%d%d", &n, &m, &l) == 3){
        kase++; num = 0;
        for(int i = 0; i < n; i++)
            scanf("%I64d", &a[i]);
        for(int i = 0; i < m; i++)
            scanf("%I64d", &b[i]);
        for(int i = 0; i < l; i++)
            scanf("%I64d", &c[i]);

        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
                ab[num++] = a[i] + b[j];
        std::sort(ab, ab+num);
        std::sort(c, c+l);

        scanf("%d", &s);
        printf("Case %d:
", kase);
        while(s--){
            scanf("%I64d", &x);
            if(x < ab[0] + c[0] || x > ab[num-1] + c[l-1])
                printf("NO
");
            else{
                int flag = 0;
                for(int i = 0; i < l; i++){
                    __int64 p = x - c[i];
                    if(_search(p)){
                        printf("YES
");
                        flag = 1;
                        break;
                    }
                }
                if(!flag)
                    printf("NO
");
            }
        }
    }
    return 0;
}