PAT甲级(Advanced Level) 1001 A+B Format

1001 A+B Format

分数 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 Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −106≤a,b≤106. 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

题意概述 

计算 a+b 并以标准格式输出总和 —— 即，数字必须用逗号分隔成三位一组（除非少于四位数字）。

题解

因为标准格式是从个位数字开始，每三个为一组（这个是个坑 一开始看题没发现这个 从正着写只拿了14分），所以要将计算后的数转化为字符串，并且反着模拟

最后将结果再反着输出

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl '\n'
ll a,b;
ll ans;
int main()
{ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>a>>b;ans=a+b;string s=to_string(ans);reverse(s.begin(),s.end());string ss="";int op=0;for(int i=0;i<s.size();i++){if(s[i]>='0'&&s[i]<='9')//判断是否是数字 不能算符号{op++;}ss+=s[i];if(op==3 && (s[i+1]>='0'&&s[i+1]<='9'))//达到一组条件后 且 下一个是数字而不是符号{ss+=',';op=0;}}reverse(ss.begin(),ss.end());cout<<ss<<endl;return 0;
}