Problem Description
The Army of Coin-tossing Monkeys (ACM) is in the business of producing randomness. Good random numbers are important for many applications, such as cryptography, online gambling, randomized algorithms and panic attempts at solutions in the last few seconds of programming competitions.
Recently, one of the best monkeys has had to retire. However, before he left, he invented a new, cheaper way to generate randomness compared to directly using the randomness generated by coin-tossing monkeys. The method starts by taking an undirected graph with 2n nodes labelled 0, 1, …, 2n - 1. To generate k random n-bit numbers, they will let the monkeys toss n coins to decide where on the graph to start. This node number is the first number output. The monkeys will then pick a random edge from this node, and jump to the node that this edge connects to. This new node will be the second random number output. They will then select a random edge from this node (possibly back to the node they arrived from in the last step), follow it and output the number of the node they landed on. This walk will continue until k numbers have been output.
During experiments, the ACM has noticed that different graphs give different output distributions, some of them not very random. So, they have asked for your help testing the graphs to see if the randomness is of good enough quality to sell.
They consider a graph good if, for each of the n bits in each of the k numbers generated, the probability that this bit is output as 1 is greater than 25% and smaller than 75%.
Input
The input will consist of several data sets. Each set will start with a line consisting of three numbers k, n, e separated by single spaces, where k is the number of n-bit numbers to be generated and e is the number of edges in the graph (1 ≤ k ≤ 100, 1 ≤ n ≤ 10 and 1 ≤ e ≤ 2000). The next e lines will consist of two space-separated integers v1, v2 where 0 ≤ v1, v2 < 2n and v1 ≠ v2. Edges are undirected and each node is guaranteed to have at least one edge. There may be multiple edges between the same pair of nodes.
The last test case will be followed by a line with k = n = e = 0, which should not be processed.
Output
For each input case, output a single line consisting of the word Yes if the graph is good, and No otherwise.
Sample Input
10 2 3
0 3
1 3
2 3
5 2 4
0 1
0 3
1 2
2 3
0 0 0
Sample Output
No
Yes
int k = 1;
00007FF6C5061575 mov dword ptr [k],1
int t=(k = 2)*++k;
00007FF6C506157D mov dword ptr [k],2
00007FF6C5061585 mov eax,dword ptr [k]
00007FF6C5061589 inc eax
00007FF6C506158B mov dword ptr [k],eax
00007FF6C506158F mov eax,dword ptr [k]
00007FF6C5061593 imul eax,dword ptr [k]
00007FF6C5061598 mov dword ptr [t],eax
从汇编来看，vs编译器是先执行了k=2，然后++k，最后计算k*k
因为k=2返回的是左值而不是右值，所以这样是合理的
比较早的编译器在定义k的时候加上volatile会出现返回右值的情况，这种情况下结果会是6
举报