PAT甲级1001 A+B Format (20分)

最新推荐文章于 2022-05-05 21:06:03 发布

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最新推荐文章于 2022-05-05 21:06:03 发布

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文章标签：算法字符串数据结构

本文链接：https://blog.youkuaiyun.com/weixin_43795683/article/details/104251408

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本文详细解析了PAT甲级1001A+BFormat题目的解题思路，介绍了一种使用C++实现的解决方案，通过字符串处理和逆序遍历技巧，实现了数字的千位分隔符格式输出。

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#PAT甲级1001 A+B Format (20分)

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 pair of integers a and b
. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

第一题解题思路：输入int 型变量a与b，将a与b相加赋给c，如果c小于0则输出一个负号，将结果利用to-string函数变为字符串形式s，然后将该字符串逆置，用s1字符串存储答案，对于s的逆序进行遍历，每隔三个字符输出一个‘，’，若最后一个字符序号恰好为3的整数倍，则不用输出‘，’，最后将s1逆置回来，输出s1。

AC代码附上：

#include <iostream>
#include <algorithm>
using namespace std;

int main(){
    int a,b,c;
    string s,s1="";
    cin>>a>>b;
    c=a+b;
    if(c<0){
        cout<<"-";
        c=-c;
    }
    s=to_string(c);
    reverse(s.begin(),s.end());
    for(int i=0;i<s.length();i++){
        s1+=s[i];
       if((i+1)%3==0&&(i!=s.length()-1)){
           s1+=',';
       }


    }
    reverse(s1.begin(),s1.end());
    cout<<s1;
    return 0;
}