1001 A+B Format (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 Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −10
​6
​​ ≤a,b≤10
​6
​​ . 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

题解

对比多种方法后，参考了柳神的思路，因为tostring有点小问题，学习了下stringstream；简单理解为，传值转换，输出
（i+1)%3==len%3&&i!=len-1这个公式不是很理解，但是i%3==0会有两个测试点无法通过

#include <iostream>
#include<string>
#include<sstream>
using namespace std;
int main()
{
int a;
int b;
string result;
stringstream c;
cin>>a>>b;
c<<a+b;
c>>result;
int len=result.length();
for(int i=0;i<len;i++){
cout<<result[i];
if (result[i] == '-') continue;
if((i+1)%3==len%3&&i!=len-1) cout<<",";
}
return 0;
}

思路2：
柳神的算法有点看不懂，所以自己想了个，（len-（i+1))%3==0,从末尾开始计数，满3就加逗号`
二刷更新，上述表达式不太好懂，可以写成

（len-1-i）%3==0

字符串长度减去正序遍历成为倒序遍历。

#include <iostream>
#include<string>
#include<sstream>
using namespace std;
int main()
{
int a;
int b;
string result;
stringstream c;
cin>>a>>b;
c<<a+b;
c>>result;
int len=result.length();
for(int i=0;i<len;i++){
cout<<result[i];
if (result[i] == '-') continue;
// if((len-(i+1))%3==0&&i!=len-1)
f((len-1-i)%3==0&&i!=len-1)
cout<<",";
}
return 0;
}

`

最后

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