hdu5650 so easy 组合数_hdu 5650-优快云博客

本文链接：https://blog.youkuaiyun.com/EventQueue/article/details/50990174

探讨了给定整数数组中所有子集的异或运算结果的总和计算问题，通过分析元素出现频率，得出只有当数组长度为1时结果才不为0的结论。

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so easy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 161 Accepted Submission(s): 130

Problem Description

Given an array with

n

integers, assume

f(S) as the result of executing xor operation among all the elements of set

S . e.g. if

S={1,2,3} then

f(S)=0 .

your task is: calculate xor of all

f(s) , here

s⊆S .

Input

This problem has multi test cases. First line contains a single integer

T(T≤20) which represents the number of test cases.
For each test case, the first line contains a single integer number

n(1≤n≤1,000) that represents the size of the given set. then the following line consists of

n different integer numbers indicate elements(

≤109 ) of the given set.

Output

For each test case, print a single integer as the answer.

Sample Input

Sample Output

  
  
   
   0

In the sample，$S = \{1, 2, 3\}$, subsets of $S$ are: $\varnothing$, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}

Source

BestCoder Round #77 (div.2)

因为偶数个相同的数异或得0，奇数个相同的数异或得自身。

发现一个规律，每个元素在所有子集中出现的次数都一样，而且都是2^(n-1)，比如样例

是C(1, 1) + C(2, 1) + C(2, 2) = 2 ^ (3 - 1)

当n = 4的时候再推一推发现为C(1, 1) + C(3, 1) + C(3, 2) +C(3, 3) = 2^(4 - 1) = 8

其实就是个杨辉三角的关系

1 2 ^ (1 - 1) = 1

1 1 2^(2 -1) = 2

1 2 1 2^(3 - 1) = 4

1 3 3 1 2 ^ (4 - 1) = 8

.....................

当n为1的时候，也就是只有一个元素的时候输出自身，其他情况所有元素在所有子集中出现的次数都是偶数，最后异或的结果肯定为0

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>

using namespace std;

long long a[1005];

int main()
{
	int T, n;
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		for (int i = 0; i < n; i++) {
			scanf("%I64d", &a[i]);
		}
		if (n == 1) {
			printf("%I64d\n", a[0]);
		}
		else {
			puts("0");
		}
	}
	return 0;
}