zoukankan      html  css  js  c++  java
  • 1001 A+B Format (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

    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
    //先求和再将和存储到数组中,倒叙三位一组输出 
    #include<cstdio>
    int num[10];
    int main(){
        int a,b;
        scanf("%d%d",&a,&b);
        int sum = a + b;
        if(sum < 0){
            printf("-");
            sum = -sum;
        }
        int len = 0; 
        if(sum == 0) num[len++] = 0; //可以用do while()循环就不用对len=0进行特殊处理 
        while(sum){
            num[len++] = sum % 10;
            sum /= 10;
        }
        for(int i = len - 1; i >= 0; i--){
            printf("%d",num[i]);
            if(i > 0 && i % 3 == 0) printf(",");
        }
        return 0;
    }
    //printf输出格式中,%3d输出三位整数,位数不足高位补空格 %03位数不足补0 
    #include<cstdio>
    int num[10];
    int main(){
        int a,b;
        scanf("%d%d",&a,&b);
        int sum = a + b;
        if(sum < 0){
            printf("-");
            sum = -sum;
        }
       if(sum >= 1000000)  printf("%d,%03d,%03d",sum/1000000,sum%1000000/1000,sum%1000); //第二个数要mod1000000 且等号不能少
       else if(sum > 1000) printf("%d,%03d",sum/1000,sum%1000); //1000通过 
       else {
           printf("%d",sum);
       }
        return 0;
    }

     2019.7.7

    #include<stdio.h>
    
    int main()
    {
        int a,b;
        int arr[10];
        int sum;
        scanf("%d%d",&a,&b);
        sum = a + b;
        int len = 0;
        if(sum < 0)
        {
            printf("-");
            sum = -sum;
        }
        if(sum == 0)
        {
            printf("0
    ");
        }
        while(sum)
        {
            arr[len++] = sum % 10;
             sum /= 10;
        }
        for(int i = len - 1; i >= 0; i--)
        {
            printf("%d",arr[i]);
            if(i % 3 == 0 && i != 0)
            {
                printf(",");
            }
        }
        return 0;
    }
  • 相关阅读:
    庄家试盘的K线形态
    股票基本知识入门提纲
    我与猫
    夜雨不眠时
    快速排序
    由float转std::string的方法
    BugFree + EasyPHP在Windows平台搭建步骤详解
    安装VS2008的时候Windows Mobile 5.0 SDK R2 for pocket pc错误解决方案
    收集WCF文章
    linq to ef(相当于sql中in的用法)查询语句
  • 原文地址:https://www.cnblogs.com/wanghao-boke/p/9384611.html
Copyright © 2011-2022 走看看