google code jam 2011 Qualification Round 资格赛 Problem C. Candy Splitting

Problem

Sean and Patrick are brothers who just got a nice bag of candy from their parents. Each piece of candy has some positive integer value, and the children want to divide the candy between them. First, Sean will split the candy into two piles, and choose one to give to Patrick. Then Patrick will try to calculate the value of each pile, where the value of a pile is the sum of the values of all pieces of candy in that pile; if he decides the piles don't have equal value, he will start crying.

Unfortunately, Patrick is very young and doesn't know how to add properly. He almost knows how to add numbers in binary; but when he adds two 1s together, he always forgets to carry the remainder to the next bit. For example, if he wants to sum 12 (1100 in binary) and 5 (101 in binary), he will add the two rightmost bits correctly, but in the third bit he will forget to carry the remainder to the next bit:

  1100+ 0101------  1001

So after adding the last bit without the carry from the third bit, the final result is 9 (1001 in binary). Here are some other examples of Patrick's math skills:

5 + 4 = 17 + 9 = 1450 + 10 = 56

Sean is very good at adding, and he wants to take as much value as he can without causing his little brother to cry. If it's possible, he will split the bag of candy into two non-empty piles such that Patrick thinks that both have the same value. Given the values of all pieces of candy in the bag, we would like to know if this is possible; and, if it's possible, determine the maximum possible value of Sean's pile.

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case is described in two lines. The first line contains a single integer N, denoting the number of candies in the bag. The next line contains the N integers C_i separated by single spaces, which denote the value of each piece of candy in the bag.

Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1). If it is impossible for Sean to keep Patrick from crying, y should be the word "NO". Otherwise, y should be the value of the pile of candies that Sean will keep.

Limits

1 ≤ T ≤ 100.
1 ≤ C_i ≤ 10⁶.

Small dataset

2 ≤ N ≤ 15.

Large dataset

2 ≤ N ≤ 1000.

Sample

Input
 
Output
  2
5
1 2 3 4 5
3
3 5 6
Case #1: NO
Case #2: 11



solution: 题目需要哥哥分配这堆数字，分配成两堆不为空的数字，并且分别对两堆数字里的所有数字进行异或运算，如果两堆的异或相等，则为一种分配方案，而现在要在可能的分配方案中使弟弟的数字和为最大，如果无法分配则弟弟为哭泣，输出NO，上面的例子中

3 5 6 可以发现5 ^ 6 == 3 所以这是一种分配的方案，而且 5 + 6 == 11此时弟弟的总和也是最大的。

 

analysis： 什么情况下可以异或的等分？

可以知道 两堆中的异或是相等的，而相等的数据异或之后为0，那么只要对所有的数据进行异或，如果为0 则说明可以等分，如果不为0 则不可能等分输出NO

                 当所有数据异或为0时，弟弟如何可以得到最大值？

同样，如果为0，可以知道，无论怎么将数据分成两堆，两堆的异或之后都是相等的，题目要求两堆的数据都不为空，那么只要让哥哥得到那个最小的数据，其余的N - 1个数据全部给弟弟，弟弟就可以得到最大值

下面给出c++的代码，时间复杂度为O（n ）（用例的那层循环不算在内）

#include<iostream>#include<fstream>using namespace std;int main(){int T;int N;int result = 0;int count = 0;ofstream out("A-large-practice.txt");cin>>T;while(count < T){cin>>N;result = 0;int *test = new int[N];int flag = 0;int sum = 0;int min = ~0 ^ (1 << 31);//使得min初始化为int里最大的数据for(int i = 0; i < N; i++){cin>>test[i];flag ^= test[i];sum += test[i];if(min > test[i])min = test[i];}result = sum - min;count++;if( flag != 0)out<<"Case #"<<count<<": "<<"NO"<<endl;if(flag == 0)out<<"Case #"<<count<<": "<<result<<endl;delete [] test;}out.close();return 0;}