PAT_A1120#Friend Numbers
Source:
Description:
Two integers are called "friend numbers" if they share the same sum of their digits, and the sum is their "friend ID". For example, 123 and 51 are friend numbers since 1+2+3 = 5+1 = 6, and 6 is their friend ID. Given some numbers, you are supposed to count the number of different frind ID's among them.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N. Then N positive integers are given in the next line, separated by spaces. All the numbers are less than 1.
Output Specification:
For each case, print in the first line the number of different frind ID's among the given integers. Then in the second line, output the friend ID's in increasing order. The numbers must be separated by exactly one space and there must be no extra space at the end of the line.
Sample Input:
8
123 899 51 998 27 33 36 12
Sample Output:
4
3 6 9 26
Keys:
- 简单模拟
Code:
/*
Data: 2019-08-14 16:21:28
Problem: PAT_A1120#Friend Numbers
AC: 13:24 题目大意:
数字的各位和称为朋友ID,求有多少个朋友ID
*/
#include<cstdio>
#include<set>
#include<string>
#include<iostream>
using namespace std; int main()
{
#ifdef ONLINE_JUDGE
#else
freopen("Test.txt", "r", stdin);
#endif // ONLINE_JUDGE int n,cnt=;
string s;
set<int> st;
scanf("%d", &n);
for(int i=; i<n; i++)
{
cin >> s;
cnt=;
for(int j=; j<s.size(); j++)
cnt += s[j]-'';
st.insert(cnt);
}
cnt=;
printf("%d\n", st.size());
for(auto it=st.begin(); it!=st.end(); it++)
printf("%d%c", *it,++cnt==st.size()?'\n':' '); return ;
}
PAT_A1120#Friend Numbers的更多相关文章
- Java 位运算2-LeetCode 201 Bitwise AND of Numbers Range
在Java位运算总结-leetcode题目博文中总结了Java提供的按位运算操作符,今天又碰到LeetCode中一道按位操作的题目 Given a range [m, n] where 0 <= ...
- POJ 2739. Sum of Consecutive Prime Numbers
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20050 ...
- [LeetCode] Add Two Numbers II 两个数字相加之二
You are given two linked lists representing two non-negative numbers. The most significant digit com ...
- [LeetCode] Maximum XOR of Two Numbers in an Array 数组中异或值最大的两个数字
Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum re ...
- [LeetCode] Count Numbers with Unique Digits 计算各位不相同的数字个数
Given a non-negative integer n, count all numbers with unique digits, x, where 0 ≤ x < 10n. Examp ...
- [LeetCode] Bitwise AND of Numbers Range 数字范围位相与
Given a range [m, n] where 0 <= m <= n <= 2147483647, return the bitwise AND of all numbers ...
- [LeetCode] Valid Phone Numbers 验证电话号码
Given a text file file.txt that contains list of phone numbers (one per line), write a one liner bas ...
- [LeetCode] Consecutive Numbers 连续的数字
Write a SQL query to find all numbers that appear at least three times consecutively. +----+-----+ | ...
- [LeetCode] Compare Version Numbers 版本比较
Compare two version numbers version1 and version1.If version1 > version2 return 1, if version1 &l ...
随机推荐
- [bzoj4025]二分图_LCT
二分图 bzoj-4025 题目大意:给定一个n个节点的图,m条边,每条边有一个产生时间和一个删除时间,询问所有时间点是否是连通图. 注释:$1\le n\le 10^5$,$1\le m\le 2\ ...
- 在CentOS VPS上源代码安装高版本号git
背景:个别软件在国内下载非常慢,在vps下载就非常快. 可是下载好后的文件通过scp弄出来的时候又非常慢,所以想通过在vps里安装git,通过gitlab或oschina来进行中转.但遗憾的是,上传到 ...
- linux下oracle11G DG搭建(二):环绕主库搭建操作
linux下oracle11G DG搭建(二):环绕主库搭建操作 环境 名称 主库 备库 主机名 bjsrv shsrv 软件版本号 RedHat Enterprise5.5.Oracle 11g 1 ...
- DirectX11 学习笔记8 - 最简单的光照
在上一个列子的基础上加了一个地面.这个地面是光照效果生成的. 看图: 先说明: 光照 须要重写一个 lightshader 就是光照的渲染器 // Define the input layout D ...
- HDU 3718 Similarity(KM最大匹配)
HDU 3718 Similarity 题目链接 题意:给定一个标准答案字符串,然后以下每一行给一个串.要求把字符一种相应一种,要求匹配尽量多 思路:显然的KM最大匹配问题,位置相应的字符连边权值+1 ...
- oc43--野指针和空指针
// // main.m // 野指针和空指针 #import <Foundation/Foundation.h> #import "Person.h" int mai ...
- LA6878
区间dp dp[i][j]存i->j区间的所有取值 然后枚举分割点,枚举两个存的值,分别运算存储. 看见这种不确定分割顺序,两个区间合并的情况,就要用区间dp. #include<bits ...
- IP Address
http://poj.org/problem?id=2105 #include<stdio.h> #include<string.h> int main() { ]; ] = ...
- android 可拖动控件 ontouchevent
首先附上文章的转载内容的链接: 学习android 可拖动事件首先需要对android的屏幕和touchevent参数建立一个详细的知识结构. 1.android坐标系统 一.首先明确一下 andro ...
- java中强制类型转换时,高位数截取成低位数的方法
/** * 强制类型转换中的补码.反码.原码一搞清楚 */ int b=233;//正整数强转 System.out.println((byte)b); //负数:原码的绝对值取反再加一 符号为不变 ...