In [23]: def reverse1(): ...: s=input("请输入需要反转的内容:") ...: return s[::-1] ...: ...: reverse1() 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 请输入需要反转的内容:你是一个小南瓜 Out[23]: '瓜南小个一是你' 参考stackflow上的答案。 原理是:This is extended sl...
Initializing the JIT engine with a stack: >>> jitter=machine.jitter(loc_db,jit_type='python') >>> jitter.init_stack() Add the shellcode in an arbitrary memory location: >>> run_addr=0x40000000>>>frommiasm.jitter.cstsimportPAGE_READ,PAGE_WRITE>>> jitter.vm.add_memory_page(run_addr,...
(BTW, I set my shell up to alias pc (Python calculator) for rpn.py to minimize typing and be a bit more like dc. But I'll use rpn.py in the examples below.)# Add two numbers. The stack is printed after all commands are run. $ rpn.py 4 5 + 9 # Do the same thing, but ...
frompwnimport*fromoperatorimportmethodcallerdefseed(x):return(((x>>6)&(x>>2)&1)==0)|((2*x)&0xff)classObj(object):s=[]defa_func(self):res=""foriinrange(10):res+=str(hex(obj.s[i]))[2:]log.critical(res)defb_func(self):forjinrange(10):self.s[j]-=1defc_func(self):f...
Leetcode: Reverse Words in a String 题目: Given an input string, reverse the string word by word. For example, Given s = “the sky is blue”, return “blue is sky the”. 思路一: 先休整下给定的字符串,去掉其中的多于一个的空格,就是使所有单词之间的空格都变成一个,然后去掉最后面的空格,...
stack.push(stack.pop() * stack.pop()); break; case"/": intn1 = stack.pop(), n2 = stack.pop(); stack.push(n2 / n1); break; default: stack.push(Integer.parseInt(a[i])); } } returnstack.pop(); } Python: 1 2 3 4
stack=[]foriinrange(0,len(tokens)):iftokens[i] !='+'andtokens[i] !='-'andtokens[i] !='*'andtokens[i] !='/': stack.append(int(tokens[i]))else: a=stack.pop() b=stack.pop()iftokens[i] =='+': stack.append(a+b)iftokens[i] =='-': ...
stack=[]# 列表实现栈的 先入后出功能count=0whilecur:count=count+1stack.append(cur)# 入栈cur=cur.nextifcount%k==0:# 已存储k个元素count=0whilestack:# 出栈倒换tmpresult.next=stack.pop()tmpresult=tmpresult.next# 处理 stack 中剩余元素whilestack:tmpresult.next=stack.pop(0)#tmpresult=tmpres...
结合[a-gA-G] 的范围得出结果 11111suctf_ACEG_test 继续调试while ( 1 ) { v53 = *v52; // 取了一个字符 v52 = "test" v53=0x74 v54 = Buf2[0]; // 字符串长度 Buf[0] = 4 v54 = 4 if ( !(v54 & *(*(sub_1400023D0(&Buf2[1]) + 24) + 2 * v53)) )// *(sub_...
This post will discuss how to reverse a string using the stack data structure in C/C++, Java, and Python using explicit stack and call stack.