PAT甲级 1001

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 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



两种思路，第一种是今天写的，将结果预处理好存在字符串里直接输出

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int main()
{
    long long a,b,c;
    char s[100];
    while(scanf("%lld%lld",&a,&b)!=EOF)
    {
        c=a+b;
        int flag=0;
        if(c<0)
        {
            flag=1;
            c=0-c;
        }
        int l=0;
        int g,w=0;
        while(c)
        {
            g=c%10;
            c/=10;
            s[l++]=g+'0';
            w++;
            if(w%3==0)s[l++]=',';
        }
        if(a+b==0)cout << "0" <<endl;
        else
        {
            if(flag)cout <<"-";
            if(s[l-1]!=',')cout << s[l-1];
            for(int i=l-2;i>=0;i--)
                cout << s[i];
            cout <<endl;
        }
    }
}

另一种是去年写的，一边输出一边加逗号，翻看以前的代码被自己秀到了。。。

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    int a,b,c,d=1;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        a+=b;
        c=0;
        d=1;
        if(a<0){printf("-"); a=0-a;}
        b=a;
        while(a/=10)
        {
            c++;
            d*=10;
        }
        c++;
        for(;c>0;)
        {
            printf("%d",b/d);
            b%=d;
            d/=10;
            c--;
            if(!(c%3)&&c)printf(",");
        }
        printf("
");
    }
    return 0;
}