HDU2141：Can you find it?(二分)

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：分别有a,b,c三个数组，给出一个数，看这个数能不能再a,b,c,里面分别找出一个数求和得到

思路：还真是想了集中方法，一开始也想到了二分，但是二分也还是有很多技巧的，不同的二分有不同的时间消耗。。。

#include <stdio.h>
#include <algorithm>
#include <string.h>
using namespace std;

__int64 a[505],b[505],c[505],la,lb,lc;
__int64 sum[505*505],len;

int bin(__int64 x)
{
    __int64 l = 0,r = len-1;
    while(l<=r)
    {
        __int64 mid = (l+r)>>1;
        if(sum[mid] == x)
            return 1;
        else if(sum[mid]>x)
            r = mid-1;
        else
            l = mid+1;
    }
    return 0;
}

int main()
{
    __int64 m,s,cas = 1,i,j,k;
    while(~scanf("%I64d%I64d%I64d",&la,&lb,&lc))
    {
        for(i = 0; i<la; i++)
            scanf("%I64d",&a[i]);
        for(i = 0; i<lb; i++)
            scanf("%I64d",&b[i]);
        for(i = 0; i<lc; i++)
            scanf("%I64d",&c[i]);
        sort(c,c+lc);
        len = 0;
        for(i = 0; i<la; i++)
        {
            for(j = 0; j<lb; j++)
            {
               sum[len++] = a[i]+b[j];
            }
        }
        sort(sum,sum+len);
        len = unique(sum,sum+len)-sum;
        scanf("%I64d",&m);
        printf("Case %I64d:\n",cas++);
        while(m--)
        {
            scanf("%I64d",&s);
            if(s>sum[len-1]+c[lc-1] || s<sum[0]+c[0])
            {
                printf("NO\n");
                continue;
            }
            __int64 flag = 0;
            for(i = 0;i<lc;i++)
            {
                int kk = s-c[i];
                if(bin(kk))
                {
                    flag = 1;
                    break;
                }
            }
            if(flag)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }

    return 0;
}