hdu5650 so easy 组合数

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 ndifferent
integer numbers indicate elements(≤109)
of the given set.

Output

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

Sample Input

1
3
1  2  3

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;
}