A - Libra

题目链接:https://abc083.contest.atcoder.jp/tasks/abc083_a


Time limit : 2sec / Memory limit : 256MB

Score : 100 points

Problem Statement

A balance scale tips to the left if L>R, where L is the total weight of the masses on the left pan and R is the total weight of the masses on the right pan. Similarly, it balances if L=R, and tips to the right if L<R.

Takahashi placed a mass of weight A and a mass of weight B on the left pan of a balance scale, and placed a mass of weight C and a mass of weight D on the right pan.

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.

Constraints

  • 1≤A,B,C,D≤10
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print Left if the balance scale tips to the left; print Balanced if it balances; print Right if it tips to the right.


Sample Input 1

Copy
3 8 7 1

Sample Output 1

Copy
Left

The total weight of the masses on the left pan is 11, and the total weight of the masses on the right pan is 8. Since 11>8, we should print Left.


Sample Input 2

Copy
3 4 5 2

Sample Output 2

Copy
Balanced

The total weight of the masses on the left pan is 7, and the total weight of the masses on the right pan is 7. Since 7=7, we should print Balanced.


Sample Input 3

Copy
1 7 6 4

Sample Output 3

Copy
Right

The total weight of the masses on the left pan is 8, and the total weight of the masses on the right pan is 10. Since 8<10, we should print Right.

 #include <iostream>
using namespace std;
int N,A,B;
int test(int x)
{
int sum=;
while(x){
sum+=x%;
x/=;
}
if(sum<=B&&sum>=A) return ;
else return ;
}
int main()
{ while(cin>>N>>A>>B){
int s=;
for(int i=;i<=N;i++){
if(test(i)) s+=i;
}
cout<<s<<endl;
}
return ;
}

B - Some Sums

题目链接:https://abc083.contest.atcoder.jp/tasks/abc083_b


Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

Find the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).

Constraints

  • 1≤N≤104
  • 1≤AB≤36
  • All input values are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the sum of the integers between 1 and N (inclusive), whose sum of digits written in base 10 is between A and B (inclusive).


Sample Input 1

Copy
20 2 5

Sample Output 1

Copy
84

Among the integers not greater than 20, the ones whose sums of digits are between 2 and 5, are: 2,3,4,5,11,12,13,14 and 20. We should print the sum of these, 84.


Sample Input 2

Copy
10 1 2

Sample Output 2

Copy
13

Sample Input 3

Copy
100 4 16

Sample Output 3

Copy
4554
 #include <iostream>
using namespace std;
int N,A,B;
int test(int x)
{
int sum=;
while(x){
sum+=x%;
x/=;
}
if(sum<=B&&sum>=A) return ;
else return ;
}
int main()
{ while(cin>>N>>A>>B){
int s=;
for(int i=;i<=N;i++){
if(test(i)) s+=i;
}
cout<<s<<endl;
}
return ;
}

AtCoder Beginner Contest 083 (AB)的更多相关文章

  1. AtCoder Beginner Contest 084(AB)

    A - New Year 题目链接:https://abc084.contest.atcoder.jp/tasks/abc084_a Time limit : 2sec / Memory limit  ...

  2. AtCoder Beginner Contest 254(D-E)

    Tasks - AtCoder Beginner Contest 254 D - Together Square 题意: 给定一个N,找出所有不超过N的 ( i , j ),使得( i * j )是一 ...

  3. AtCoder Beginner Contest 086 (ABCD)

    A - Product 题目链接:https://abc086.contest.atcoder.jp/tasks/abc086_a Time limit : 2sec / Memory limit : ...

  4. AtCoder Beginner Contest 085(ABCD)

    A - Already 2018 题目链接:https://abc085.contest.atcoder.jp/tasks/abc085_a Time limit : 2sec / Memory li ...

  5. AtCoder Beginner Contest 264(D-E)

    D - "redocta".swap(i,i+1) 题意: 给一个字符串,每次交换相邻两个字符,问最少多少次变成"atcoder" 题解: 从左到右依次模拟 # ...

  6. Atcoder Beginner Contest 155E(DP)

    #definde HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ]; int main(){ ios: ...

  7. Atcoder Beginner Contest 156E(隔板法,组合数学)

    #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; ; ; long long fac[N] ...

  8. Atcoder Beginner Contest 121D(异或公式)

    #include<bits/stdc++.h>using namespace std;int main(){    long long a,b;    cin>>a>&g ...

  9. Atcoder Beginner Contest 155D(二分,尺取法,细节模拟)

    二分,尺取法,细节模拟,尤其是要注意a[i]被计算到和a[i]成对的a[j]里时 #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> ...

随机推荐

  1. MongoDB pymongo模块 删除数据

    使用user集合,删除user集合的数据 import pymongo mongo_client = pymongo.MongoClient( host='192.168.0.112', port=2 ...

  2. CentOS安装Yarn只需两步就搞定

    Yarn 是一个依赖管理工具.它能够管理你的代码,并与全世界的开发者分享代码.Yarn 是高效.安全和可靠的,你完全可以安心使用.代码是通过包(有时也被称为组件). 在每一个包中会定义一个 packa ...

  3. laravel用crud修改产品items-新建resource controller和routing

    前面我们创建了laravel简单的items产品api,但是需要在数据库添加,如何在网页上直接添加呢?我们可以用view来操作crud(增加Create.读取查询Retrieve.更新Update和删 ...

  4. 20165236 实验二 《Java面向对象程序设计》实验报告

    20165236 实验二<Java面向对象程序设计>实验报告 姓名:郭金涛       学号:20165236      课程:Java程序设计 指导老师:娄嘉鹏       实验时间:2 ...

  5. 网络基础之2——TCP/IP参考模型

    本内容主要来源于<看透Spring MVC源码分析与实践——韩路彪>一书 BS结构网络传输的分解方式有两种: 1.OSI参考模型. 2.TCP/IP参考模型. OSI和TCP/IP分层模型 ...

  6. [py]python写一个通讯录step by step V3.0

    python写一个通讯录step by step V3.0 参考: http://blog.51cto.com/lovelace/1631831 更新功能: 数据库进行数据存入和读取操作 字典配合函数 ...

  7. 【Linux】-NO.9.Linux.5.Nexus.1.001-【CentOS 7 Install Nexus 3.3】-

    1.0.0 Summary Tittle:[Linux]-NO.9.Linux.5.Nexus.1.001-[CentOS 7 Install Nexus 3.3]- Style:Linux Seri ...

  8. AOP 入门

    1,源码 Application.java package com.bf; import org.springframework.context.ApplicationContext; import ...

  9. Mac 安装HTMLTestRunner模块

    1.下载HTMLTestRunner.py文件 下载地址http://tungwaiyip.info/software/HTMLTestRunner.html 将下载的文件放在lib下: /Libra ...

  10. shell实现SSH自动登陆【转】

    前言 公司开发使用docker,每次登陆自己开发机总要输入 ssh user_name@ip_string,然后再确认输入password,手快了还经常会输错.作为一个懒人,肯定要找一个取巧的方式,查 ...