Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Specification: Each input file contains one test case. Each case contains a pa…

AC代码 #include <cstdio> #include <algorithm> using namespace std; const int maxn = 11; int main() { #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif // ONLINE_JUDGE long long a, b; scanf("%lld %lld",…

题目链接:传送门 题目简述: 1. 给定两个整数值a,b: 2.范围-1000000 <= a, b <= 1000000: 3.按指定格式输出结果 例:-100000 9 输出: -99,991 解题思路: 1.明确范围     a+b在正负两百万范围内, 32位系统int类型占4字节精度够 2.明确要求: ① 输入以空格分割, 输入整数 ②结果如果数字大于4位, 需要每三位用逗号分割 ③视算法可能有需要补零的情况(我就是踩的这个坑) ④正负号提前判定, 便于后面处理 ⑤函数要以return…

Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case contains a pair of integer…

GitHub PDF 1001. A+B Format (20) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). 解题思路 将a+b的值先存在sum中,再逐个分解sum的数字,将组成sum的所有数字存到数组中,由…

题目 Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Specification: Each input file contains one test case. Each case contains a…

最新想法: 最多是七位数,而且只有一组输入,完全不用考虑算法复杂度. 直接判断是否为负,并输出符号 巧妙的地方:while循环的下一次再添加逗号.(防止出现,999,991的情况) 婼姐的方法真的很巧妙,能理解,但是不知道如果是自己该怎样想出那种方法. 所以仅仅把婼姐的代码当做学习用,还是得自己来实现啊. 参考了网上很多的博客,真的不好,直接贴代码,思路很差劲的也有.下面这是我觉得最能理解的好方法. 思路分析: 1.a=a+b,当a>1000时,用ans[i++]数组存储a%1000的值,a=a…

#include<cstdio> #include<cstring> using namespace std; char s[10]; int main() { int a,b; while(scanf("%d%d",&a,&b)==2) { memset(s,0,sizeof(s)); a=a+b; if(a==0) { printf("0\n"); continue; } int aa=a; b=0; if(a<0)…

题目: 我一开始的思路是: 用math.h中的log10函数来计算位数(不建议这么做,因为会很慢,而且会出一点别的问题): 用pow函数根据要插入分号的位置来拆分a+b成一个个数字(例如res / pow(10, len - 3)来获得千位以前的数字),从左往右依次输出,同时在对应位置输出,: 也就是说,我这里的思路是直接用数字来进行处理的,但是这样做其实非常低效而且很容易写错代码,因此我看了下柳婼的代码,换成了这个思路:先将计算结果转换为字符串后进行处理.这样就会简单许多: #include…

题目AC汇总 甲级AC PAT A1001 A+B Format (20 分) PAT A1002 A+B for Polynomials(25) PAT A1005 Spell It Right (20) PAT A1006 Sign In and Sign Out (25) PAT A1009 Product of Polynomials(25) PAT A1011 World Cup Betting(20) PAT A1012 Best Rank(25) PAT A1016 Phone B…

B1024. 科学计数法 (20) Description: 科学计数法是科学家用来表示很大或很小的数字的一种方便的方法,其满足正则表达式[+-][1-9]"."[0-9]+E[+-][0-9]+,即数字的整数部分只有1位,小数部分至少有1位,该数字及其指数部分的正负号即使对正数也必定明确给出. 现以科学计数法的格式给出实数A,请编写程序按普通数字表示法输出A,并保证所有有效位都被保留. Input: 每个输入包含1个测试用例,即一个以科学计数法表示的实数A.该数字的存储长度不超过99…

从今天起每天刷1-2题PAT甲级 第一天 A1001 A+B Format (20 分) 题目内容 Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Specification: Each input fi…

问题的提出 公司各个业务线的安装包小则几十兆.大则几百兆,使用自建的升级系统向全国百万级用户下发新版本时,流量耗费相当惊人.有时新版本仅仅改了几个 dll ,总变更量不过几十 K 而已,也要发布一个完整版本.为了降低流量费用,我们推出了补丁升级的方式:产品组将修改的 dll 单独挑选出来,加上一个配置文件压缩成包,上传到自建的升级后台:在客户端,识别到补丁包类型后,手动解压并替换各个 dll 完成安装(之前是直接启动下载好的安装包).这种方式一经推出,受到了业务线的追捧.然而在使用过程中,也发现…

PAT (Advanced Level) Practice 1001 A+B Format (20 分) 凌宸1642 题目描述: Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). 译:计算 a + b 的和,并格式化…

题目原文: Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case contains a pair of i…

[题目链接] Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case contains a pair of…

题目: Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case contains a pair of int…

题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805528788582400 Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). I…

题目链接:https://pintia.cn/problem-sets/994805342720868352/problems/994805528788582400 1001 A+B Format (20 分) Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are le…

1001 A+B Format (20)(20 分) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case…

1001 A+B Format (20)(20 分) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case…

1001 A+B Format (20)(20 分) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case…

1001. A+B Format (20) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four d…

Github 1001 题目速览 1.解题的思路过程 认真读题,题目为A+BFormat,简单的计算a+b问题,特殊在于输出的形式. 输入形式为每个输入文件包含一个测试样例,每个测试样例仅包含一对整型a与b. 数字大小范围为-1000000到1000000,确定使用整型类型表示和. 输出较为特殊,结果必须使用标准形式输出,即结果为四位数及以上的必须从最后开始每三位数字前添加一个逗号,增强结果的可读性. 问题的代码实现 采用取余的方法从最后开始将三位三位取出来. 题目结果最大值为2,000,000…

https://pintia.cn/problem-sets/994805342720868352/problems/994805528788582400 Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Inpu…

problem 1001 A+B Format (20)(20 point(s)) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test…

1001. A+B Format (20) Calculate a + b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Each input file contains one test case. Each case cont…

Github的object-oriented仓库:1001.A+BFormat(20) 1.解题的思路过程 在之前学习C语言时曾经碰到过类似的将数字转换成字符输出的情况,这道题目要求输出的数字每三个间增加一个逗号,我就想到将每一位上的数字都转换成字符,并在每输出三个字符后输出一个逗号. 计算了一下最大是7位数加上两个逗号是九位,于是我开了s[10].一开始我写了一个判断,将和为负数的都转换成正数并用了变量k做了记录方便为s[10]赋值和之后的输出. 用一个循环语句给数组s[10]赋值,因为平时练…

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Specification: Each input file contains one test case. Each case contains a pa…

pat1001 A+B Format (20 分) Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits). Input Specification: Each input file contains one test cas…