codechef july月赛 The Gray-Similar Code

http://www.codechef.com/JULY12/problems/GRAYSC/

The Gray-Similar Code

Problem code: GRAYSC

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The Gray code (see wikipedia for more details) is a well-known concept.One of its important properties is that every two adjacent numbers have exactly one different digit in their binary representation.

In this problem, we will give you n non-negative integers in a sequenceA[1..n] (0<=A[i]<2^64), such that every two adjacent integers have exactly one different digit in their binary representation, similar to the Gray code.

Your task is to check whether there exist 4 numbers A[i1], A[i2], A[i3], A[i4] (1 <= i1 < i2 < i3 < i4 <= n) out of the givenn numbers such thatA[i1] xor A[i2] xor A[i3] xor A[i4] = 0. Herexor is abitwise operation which is same as^ in C, C++, Java and xor in Pascal.

Input

First line contains one integer n (4<=n<=100000).Second line containsn space seperated non-negative integers denoting the sequenceA.

Output

Output “Yes” (quotes exclusive) if there exist four distinct indices i1, i2, i3, i4 such thatA[i1] xor A[i2] xor A[i3] xor A[i4] = 0. Otherwise, output "No" (quotes exclusive) please.

Example

Input:51 0 2 3 7Output:Yes

题意：  输入n个数  每2个相邻的数对应的二进制 只有其中的一位是不同的  问是否存在4个数 使得是个数抑或等于0

思路： 相邻的两个抑或肯定只剩下1位为1    相同为0 不同为1       所以如果数字够多的话得到的结果为2^0 2^1 2^3 .......2^64  如果数量大于2*64的话 一定会有重复的

那么2个重复的抑或等于0  所以当数量大于2*64  直接输出yes  否则暴力求解 即可

下面是标程

#include <cstdio> #include <cstring> #include <cstdlib> #include <iostream> using namespace std;   const int limit=64*2+10;   int main() { int n; unsigned long long a[limit]; cin >> n; if (n>=limit){ puts("Yes"); return 0; } for (int i=0;i<n;++i) cin >> a[i]; for (int i=0;i<n;++i) for (int j=i+1;j<n;++j) for (int k=j+1;k<n;++k) for (int l=k+1;l<n;++l) if ((a[i]^a[j]^a[k]^a[l])==0){ puts("Yes"); return 0; } puts("No"); return 0; }