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

A - Can you find it?

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

有三个vector，每个vector的元素可达500个，如果直接枚举，O(n^3)这样的复杂度肯定会超时，于是将前两个vector加起来，得到 vecG ，第三个vector是 vecC 。

显然，对于 vecC 、vecG 中的元素 vecC[i] 和 vecG[j] ，以及每个令等式成立的的 X （记为 Xp）满足以下关系：

∀i∈[0,vecC.size()), ∃j∈[0,vecG.size()), Xp−vecC[i]=vecG[j]

由公式可知，对所有给出的 X ，只要使用binary_search()判断 vecG 中是否存在 X - vecC[j] 这个值，就可以知道这个 X 是不是可以由三个容器中的值组合得到。这样一来，复杂度将为O(n^2)，来自求vecG的步骤。

#include <iostream>
#include <vector>
#include <algorithm>
#include <cstdio>
using namespace std;
int main(){

int l, n, m, t = 1;
while(cin >> l >> n >> m){
int temp;
vector<int> vecA, vecB, vecC, vecS, vecG;
for(int i = 0; i < l; i++){
scanf("%d", &temp);
vecA.push_back(temp);
}
for(int i = 0; i < n; i++){
scanf("%d", &temp);
for(int j = 0; j < l; j++){
vecB.push_back(temp + vecA[j]);
}
}
sort(vecB.begin(), vecB.end());
for(int i = 0; i < vecB.size(); i++){
if(!binary_search(vecG.begin(), vecG.end(), vecB[i]))
vecG.push_back(vecB[i]);
}
for(int i = 0; i < m; i++){
scanf("%d", &temp);
vecC.push_back(temp);
}
int s;
cin >> s;
cout << "Case " << t++ << ":" << endl;
for(int i = 0 ;i < s; i++){
scanf("%d", &temp);
int j;
for(j = 0; j < vecC.size(); j++){
if(binary_search(vecG.begin(), vecG.end(), temp - vecC[j])){
cout << "YES" << endl;
break;
}
}
if(j == vecC.size())
cout << "NO" << endl;
}
}

return 0;
}