CSU暑期集训day02_A_二分法例题

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.

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

Sample Output

Case 1:
NO
YES
NO

题目大意：

给出三组数字，要求判断指定数字能否由三组数字中各取一个相加得到。

解题思路：

先将前两组数组相加组合成一个新的数组，时间复杂度是O(n2)，然后依次枚举第三个数组中的数字，在新数组中二分查找。

代码：

#include <iostream>
#include <algorithm>

using namespace std;

int main(){
   int l,m,n;
   int coun=1;
   while(cin>>l>>n>>m){
       cout<<"Case "<<coun<<':'<<endl;
       coun++;
       int arr[501]={-1};
       int brr[250001]={-1};
       int crr[501]={-1};
       for(int i=0;i<l;i++)cin>>arr[i];
       for(int i=0;i<n;i++){
           int b;
           cin>>b;
           for(int j=0;j<l;j++){
               brr[l*i+j]=b+arr[j];
           }
       }
       sort(brr,brr+l*n);
       for(int i=0;i<m;i++){
           cin>>crr[i];
       }
       int S;
       cin>>S;
       for(int i=0;i<S;i++){
           int x;
           cin>>x;
           bool ans=false;
           for(int j=0;j<m;j++){
               if(binary_search(brr,brr+l*n,x-crr[j])){
                   ans=true;
                   break;
               }
           }
           if(ans)cout<<"YES"<<endl;
           else cout<<"NO"<<endl;
       }
   }
   return 0;
}