题解报告：hdu 2141 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

解题思路：典型的二分搜索，先将a、b两两求和并排好序，然后二分搜索是否能找到x-c[j]这个值即可。 

AC代码(358ms)：

#include<bits/stdc++.h>
using namespace std;
const int maxn=505;
int l,n,m,q,cnt=1,x,a[maxn],b,c[maxn];bool flag;
vector<int> vec;
int main(){
    while(~scanf("%d%d%d",&l,&n,&m)){
        vec.clear();
        for(int i=1;i<=l;++i)scanf("%d",&a[i]);
        for(int i=1;i<=n;++i){
            scanf("%d",&b);
            for(int j=1;j<=l;++j)
                vec.push_back(a[j]+b);
        }
        sort(vec.begin(),vec.end());
        for(int i=1;i<=m;++i)scanf("%d",&c[i]);
        scanf("%d",&q);printf("Case %d:\n",cnt++);
        while(q--){
            scanf("%d",&x);flag=false;
            for(int j=1;j<=m;++j){
                int pos=lower_bound(vec.begin(),vec.end(),x-c[j])-vec.begin();
                if(pos!=l*n&&vec[pos]==x-c[j]){flag=true;break;}
            }
            if(flag)puts("YES");
            else puts("NO");
        }
    }
    return 0;
}