PAT:A1001. A+B Format (20/20)

最新推荐文章于 2020-11-06 18:59:29 发布

原创最新推荐文章于 2020-11-06 18:59:29 发布 · 288 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT #乙级

算法同时被 2 个专栏收录

39 篇文章

订阅专栏

数据结构

36 篇文章

订阅专栏

本文介绍了一个C语言程序，该程序接受两个整数作为输入，并以标准格式输出它们的和，即每三位数字用逗号分隔。讨论了处理负数和不同位数数字的细节。

1、0需要特判；

2、开始提交的代码中没有考虑好个位数的情况；

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

#include <cstdio>
#include <cstring>

int main()
{
    int a,b,sum,i=0;
    int ass[10];
    scanf("%d %d",&a,&b);
    sum=a+b;
    if(sum<0){
        printf("-");
        sum=-sum;
    }
    do{
        ass[i++]=sum%10;
        sum/=10;
    }while(sum!=0);
    if(ass[i-1]==0){
        printf("0");
        i=-1;
    }else if(i%3==2){
        printf("%d%d,",ass[i-1],ass[i-2]);
        i-=3;
    }else if(i%3==1){
        printf("%d,",ass[i-1]);
        i-=2;
    }else{
        i--;
    }
    for(;i>=0;){
        for(int j=0;j<3;j++){
            printf("%d",ass[i]);
            i--;
        }
        if(i>1)printf(",");
    }
    return 0;
}


正确提交：
#include <cstdio>
#include <cstring>

int main()
{
    int a,b,sum,i=0;
    int ass[10];
    scanf("%d %d",&a,&b);
    sum=a+b;
    if(sum<0){
        printf("-");
        sum=-sum;
    }
    do{
        ass[i++]=sum%10;
        sum/=10;
    }while(sum!=0);
    if(ass[i-1]==0){
        printf("0");
        i=-1;
    }else if(i%3==2){
        printf("%d%d",ass[i-1],ass[i-2]);
        if(i>3)printf(",");
        i-=3;
    }else if(i%3==1){
        printf("%d",ass[i-1]);
        if(i>3)printf(",");
        i-=2;
    }else{
        i--;
    }
    for(;i>=0;){
        for(int j=0;j<3;j++){
            printf("%d",ass[i]);
            i--;
        }
        if(i>1)printf(",");
    }
    return 0;
}