Digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 77503 Accepted Submission(s): 24224
For example, consider the positive integer 24. Adding the 2 and the 4 yields a value of 6. Since 6 is a single digit, 6 is the digital root of 24. Now consider the positive integer 39. Adding the 3 and the 9 yields 12. Since 12 is not a single digit, the process must be repeated. Adding the 1 and the 2 yeilds 3, a single digit and also the digital root of 39.
/* Name: hdu--1013--Digital Roots Copyright: ©2017 日天大帝 Author: 日天大帝 Date: 22/04/17 10:34 Description: 这个题,就是特别坑 如果你一开始把所有的值设置为int型,恭喜你,你会得到一个WA 接着你大概会改成unsigned型,恭喜你,你会得到一个超时 然后你终于恍然大悟,用字符串!! */ #include<iostream> #include<string> using namespace std; int main(){ ios::sync_with_stdio(false); string str; ") { ; ; ){ int temp = sum; sum = ; while(temp){ sum += temp%; temp /= ; } } cout<<sum<<endl; } ; }
hdu--1013--Digital Roots(字符串)的更多相关文章
-
HDU 1013 Digital Roots(字符串)
Digital Roots Problem Description The digital root of a positive integer is found by summing the dig ...
-
HDU 1013 Digital Roots(to_string的具体运用)
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1013 Digital Roots Time Limit: 2000/1000 MS (Java/Othe ...
-
HDU 1013 Digital Roots【字符串,水】
Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
-
HDU 1013 Digital Roots(字符串,大数,九余数定理)
Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Tot ...
-
HDU 1013.Digital Roots【模拟或数论】【8月16】
Digital Roots Problem Description The digital root of a positive integer is found by summing the dig ...
-
HDU 1013 Digital Roots 题解
Problem Description The digital root of a positive integer is found by summing the digits of the int ...
-
hdu 1013 Digital Roots
#include <stdio.h> int main(void) { int m,i;char n[10000]; while(scanf("%s",&n)= ...
-
HDU OJ Digital Roots 题目1013
/*Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Other ...
-
HDU - 1310 - Digital Roots
先上题目: Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Othe ...
-
杭电 1013 Digital Roots
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1013 反思:思路很简单,但是注意各位数加起来等于10的情况以及输入0的时候结束程序该怎么去表达 #in ...
随机推荐
-
[Bootstrap]7天深入Bootstrap(1)入门准备
由于申请了一个域名,一个云主机,开始弄个人网站. 发现Bootstrap非常方便,和重要,故开始学习与分享关于Bootstrap的技术. 推个广告 个人网站:http://www.neverc.cn ...
-
给php增加gd库(转)
1.安装zlib tar zxvf zlib-1.2.3.tar.gz cd zlib-1.2.3 ./configure --prefix=/usr/local/zlib make make ins ...
-
Perl内置变量速查表
[ 文件句柄 ] $| 如果非零, 则在对当前选定的文件执行写或打印操作后强制清除缓冲区 $% 当前选中文件句柄的当前页码 $= 当前选中文件句柄的当前页面长度 $- 当前选中文件句柄的页面剩余长度 ...
-
Oracle 分析函数 ";ORA-30485: 在窗口说明中丢失 ORDER BY 表达式";
跟顺序有关的几个分析函数row_number.rank.dense_rank.lead和lag的over窗口里,都必须有order_by_clause.其他几个如:first_value.last_v ...
-
Linq中的常用方法
System.Linq System.Linq.Enumerable 类 Range Repeat Reverse Select Where Sum Zip Aggregate Count Firs ...
-
The Child and Sequence
Codeforces Round #250 (Div. 1)D:http://codeforces.com/problemset/problem/438/D 题意:给你一个序列,然后有3种操作 1x ...
-
Sphinx Makefile
# Makefile for Sphinx documentation # # You can set these variables from the command line. SPHINXOPT ...
-
Go-单元测试
文章转载地址:https://www.flysnow.org/2017/05/16/go-in-action-go-unit-test.html 什么是单元测试? 单元测试一般用来测 ...
-
B/S架构
B/S架构即浏览器和服务器架构模式.它是随着Internet技术的兴起,对C/S架构的一种变化或者改进的架构.在这种架构下,用户工作界面是通过WWW浏览器来实现,极少部分事务逻辑在前端(Browser ...
-
Jmeter - 测试 http 接口
前言: 本文主要针对http接口进行测试,使用Jmeter工具实现. Jmter工具设计之初是用于做性能测试的,它在实现对各种接口的调用方面已经做的比较成熟,因此,本次直接使用Jmeter工具来完成对 ...