CSU-ACM2017暑假集训2-二分搜索 hdu2141- Can you find it?

题目：

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. 

InputThere 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. 

OutputFor 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

           题目大意：问能否在A,B,C,三个数组内各选一个数，使得他们的和为给定的值X

           思路：先选两个数组比如A,B求其各个数相加之和，存到一个数组T中，再枚举数组C的数得到满足X-c=a+b的值，再在T数组中二分查找是否            存在X-c为A,B之和。复杂度O(n*n).

           代码：

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

int l,n,m;
int a[555],b[555],c[555];
int a_b[555*555];
int a_b_norep[555*555];
int num_ab;

void init()
{
int k,t;
k=t=0;
for(int i=0;i<l;i++)
scanf("%d",&a[i]);
for(int i=0;i<n;i++)
scanf("%d",&b[i]);
for(int i=0;i<m;i++)
scanf("%d",&c[i]);
for(int i=0;i<l;i++)
{
for(int j=0;j<n;j++)
{
a_b[k++]=a[i]+b[j];
}
}
sort(a_b,a_b+k);
a_b_norep[t++]=a_b[0];
for(int i=1;i<k;i++)
if(a_b[i]!=a_b[i-1])
a_b_norep[t++]=a_b[i];
num_ab=t;
}

int bs(int x)
{
int l,r,mid;
l=0;
r=num_ab;
while(l<=r)
{
mid=(r-l)/2+l;
if(a_b_norep[mid]==x)
return 1;
else if(a_b_norep[mid]<x)
l=mid+1;
else
r=mid-1;
}
return 0;
}

int main()
{
int t,x;
int no=0;
while(scanf("%d%d%d",&l,&n,&m)!=EOF)
{
init();
scanf("%d",&t);
printf("Case %d:\n",++no);
while(t--)
{
int flag=0;
scanf("%d",&x);
for(int i=0;i<m;i++)
{
if(bs(x-c[i]))
{
flag=1;
break;
}
}
if(flag)
printf("YES\n");
else
printf("NO\n");
}
}
return 0;
}