PAT甲级-1001. A+B Format (20)逗号分隔

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本文介绍了一个简单的C++程序，用于计算两个整数A和B的和，并将结果以标准格式输出，即每三位数字用逗号分隔。该程序考虑了正负数的情况，并通过循环实现了整数到字符串的转换及格式化。

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 digits).

Input

Each input file contains one test case. Each case contains a pair of integers a and b where -1000000 <= a, b <= 1000000. The numbers are separated by a space.

Output

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

计算A+B的值，结果的每3个数位用一个逗号分隔开。

#include<bits/stdc++.h>
using namespace std;
#define MAXN 1000010
#define INF 0xfffffff
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("F:/cb/read.txt","r",stdin);
    //freopen("F:/cb/out.txt","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int a,b,ans,i,j,cnt=0;
    bool flag=true;//正负标志
    cin>>a>>b;
    ans=a+b;
    //cout<<"ans="<<ans<<endl;
    if(ans<0) flag=false,ans=abs(ans);
    char s[MAXN],c[MAXN];
    while(ans>0)//int转字符串
    {
        s[cnt++]=char(ans%10+'0');
        ans/=10;
    }
    if(ans==0) c[0]='0';
    cnt=0;
    for(i=strlen(s)-1; i>=0; --i)
        c[cnt++]=s[i];
    //cout<<"c="<<c<<endl;
    int len=strlen(c);
    int temp=len/3;
    if(len%3==0) --temp;//逗号个数
    //cout<<"len="<<len<<endl;
    cnt=0;
    if(!flag) cout<<"-";
    for(i=0; i<len-3*temp; ++i)
        cout<<c[i];
    if(temp>0)
    {
        cout<<",";
        for(j=i; j<len; ++j)
        {
            if(cnt==3)
            {
                cnt=0;
                cout<<",";
            }
            cout<<c[j];
            ++cnt;
        }
    }
    cout<<endl;
    return 0;
}