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HDU 5650 so easy

时间:2016-04-26 22:07:18      阅读:163      评论:0      收藏:0      [点我收藏+]

so easy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617    Accepted Submission(s): 413


Problem Description
Given an array with
n技术分享
integers, assume f(S)技术分享 as the result of executing xor operation among all the elements of set S技术分享. e.g. if S={1,2,3}技术分享 then f(S)=0技术分享.

your task is: calculate xor of all f(s)技术分享, here s?S技术分享.
 

Input
This problem has multi test cases. First line contains a single integer T(T≤20)技术分享 which represents the number of test cases.
For each test case, the first line contains a single integer number n(1≤n≤1,000)技术分享 that represents the size of the given set. then the following line consists of n技术分享 different integer numbers indicate elements(≤10技术分享9技术分享技术分享) 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
 

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题意:给出N个数的集合,这个集合的所有子集合的元素异或运算后,再将这些子集合的运算结果进行异或运算,问最后结果。
做这道题需要知道的是异或运算满足交换律(a^b)^(c^d)=a^b^c^d  ,并且a^a=0;  a^0=a;只有N为1时,集合中的元素只出现奇数次,其他情况集合中的元素都是出现偶数次,所以此时结果都为0;
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int main()
{
	int t,i,j,k,l,m,n;
    scanf("%d",&t);
    while(t--)
    {
    	scanf("%d",&n);
    	scanf("%d",&m);
    	for(i=1;i<n;i++)
    	scanf("%d",&k);
    	if(n==1)
    	printf("%d\n",m);
    	else
    	printf("0\n");
	}
	return 0;
}


 

HDU 5650 so easy

原文:http://blog.csdn.net/yao1373446012/article/details/51228650

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