luis — Fri, 24 Jul 2009 02:59:00 GMT

You are to write an interpreter for a simple computer. This computer uses a processor with a small number of machine instructions. Furthermore, it is equipped with 32 byte of memory, one 8-bit accumulator (accu) and a 5-bit program counter (pc). The memory contains data as well as code, which is the usual von Neumann architecture.

The program counter holds the address of the instruction to be executed next. Each instruction has a length of 1 byte - the highest 3 bits define the type of instruction and the lowest 5 bits define an optional operand which is always a memory address (xxxxx). For instructions that don't need an operand the lowest 5 bits have no meaning (-----). Here is a list of the machine instructions and their semantics:

000xxxxx STA x store the value of the accu into memory byte x
001xxxxx LDA x load the value of memory byte x into the accu
010xxxxx BEQ x if the value of the accu is 0 load the value x into the pc
011----- NOP no operation
100----- DEC subtract 1 from the accu
101----- INC add 1 to the accu
110xxxxx JMP x load the value x into the pc
111----- HLT terminate program

In the beginning, program counter and accumulator are set to 0. After fetching an instruction but before its execution, the program counter is incremented. You can assume that programs will terminate.

Input Specification

The input file contains several test cases. Each test case specifies the contents of the memory prior to execution of the program. Byte 0 through 31 are given on separate lines in binary representation. A byte is denoted by its highest-to-lowest bits. Input is terminated by EOF.

Output Specification

For each test case, output on a line the value of the accumulator on termination in binary representation, again highest bits first.

Sample Input

Sample Output

10000111

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luis 2009-07-24 10:59 发表评论

joj 2199 the lost number 好题值得做做

luis — Sun, 19 Jul 2009 06:18:00 GMT

Status	In/Out	TIME Limit	MEMORY Limit	Submit Times	Solved Users	JUDGE TYPE
	stdin/stdout	3s	40960K	1071	235	Standard

In a interger sequence S there are N(N < 1000000) intergers, there is a initial number I(-2^31 < I <2^31), which is the minimun interger in S, and no two integers are the same. Now can you find the first lost interger L make the sequence is not consecutive. For example, S = { 1, -2, 2, 9, -1 }, then I = -2, and L = 0.

Input

For each case N is in the first line, and next N lines is the sequence S.

Output

Output L for each case in a single line.

Sample Input

Sample Output

不用走入排序的误区，

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先一�����@环，保存输入的值n[MAX],同时可找出最���的��|��
再来一�����@环以此�ؓ基准�Ҏ(gu��)��个数�q�行标记�Q�对每个出现的num,mark[num-min]=1;
然后
for(int i=0;;i++)
{if(mark[i]==0)
cout<
break;
}
----

luis 2009-07-19 14:18 发表评论

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Input Specification

Output Specification

Sample Input

Sample Output

joj 2199 the lost number 好题 值得做做

Input

Output

Sample Input

Sample Output

joj 2199 the lost number 好题值得做做