整数越界情况 1. 数组下标越界, 大于N或者小于0 2. 数字过大,可以选择取个模,或者换long long, double  我笑了 还有一个暂时没有好的解决方法的:string s:cin/输入后才能使用较大的下标:若定义后手动赋值且下标较大也会显示越界(40就算较大下标,手动狗头) 欢迎广大网友指正…

0x00      Preview Last few passage I didn't conclude some important points and a general direction of the learning route of PWN. I browser some passages and finally I found one maybe suitable and reasonable for most PWN lovers: 0x01  Integer Overflow…

Reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123, return -321 click to show spoilers. Have you thought about this? Here are some good questions to ask before coding. Bonus points for you if you have already thought throu…

Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21 题意: 给定一个10进制整数,翻转它. Solution1: directly do the simulation. Two tricky parts to be hand…

题目等级:Easy 题目描述: Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21    题意:给出一个 32 位的有符号整数,你需要将这个整数中每位上的数字进行反转. 解题思路:   本题很简单,我们给出以下两种方法.  …

题目 Given a 32-bit signed integer, reverse digits of an integer. Example1: x = 123, return 321 Example2: x = -123,return -321 Example3: x = 120, return 21 思路 问题的关键在于对溢出的判断. 思路1:用long或者long long 直接用long或者long long计算,这样就保证不会溢出,再判断反转后的数字是否溢出. 思路2:stoi()…

Given a 32-bit signed integer, reverse digits of an integer. Example 1: Input: 123 Output: 321 Example 2: Input: -123 Output: -321 Example 3: Input: 120 Output: 21 Note:Assume we are dealing with an environment which could only store integers within…

题目: 颠倒整数 将一个整数中的数字进行颠倒,当颠倒后的整数溢出时,返回 0 (标记为 32 位整数). 样例 给定 x = 123,返回 321 给定 x = -123,返回 -321 解题: 直接反转,越界处理好炒蛋 Java程序: public class Solution { /** * @param n the integer to be reversed * @return the reversed integer */ public int reverseInteger(int n…

主要内容:增加两个整数溢出检测 #include <stdio.h> #include <limits.h> int main(int argc, char *argv[]) { /* * a和b为非负整型变量,检測a+b是否会"溢出" */ // INT_MAX=2147483647 int a=123456789,b=2147483000; /* 方法一: if(a+b < 0) printf("overflow\n"); 错误原因…

整数溢出问题 Java 中的 int 用 32 位表示,正数最大值的情况,首位是 0,其他位都可以是 1(就是 2^31-1).但是如果正数过大了,例如 2^31,计算机不得不把首位变成 1,并且计算机不知道这是溢出情况,把它按照正常的方式输出了,于是就成了负的. 2^31 - 1 = 0111 1111 1111 1111 1111 1111 1111 1111 = 2147483647 2^31      = 2^31 - 1 + 1 = 1000 0000 0000 0000 0000 0…