Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4436    Accepted Submission(s): 2505 Problem Description The digital root of a positive integer is found by summing the digi…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5238    Accepted Submission(s): 2925 Problem Description The digital root of a positive integer is found by summing the digits of the integer. If…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4793    Accepted Submission(s): 2672 Problem Description The digital root of a positive integer is found by summing the digits of the integer. If…

Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5783    Accepted Submission(s): 3180 Problem Description The digital root of a positive integer is found by summing the digit…

Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits a…

Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5113    Accepted Submission(s): 2851 Problem Description The digital root of a positive integer is found by summing the digit…

Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5183    Accepted Submission(s): 2897 Problem Description The digital root of a positive integer is found by summing the digit…

Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5745    Accepted Submission(s): 3160 Problem Description The digital root of a positive integer is found by summing the digi…

Digital Roots 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:456            测试通过:162 描述 The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the d…

Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 57857 Accepted Submission(s): 18070 Problem Description The digital root of a positive integer is found by summing the digits of the i…

http://acm.njupt.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=1028 Digital Roots 时间限制(普通/Java):1000MS/3000MS          运行内存限制:65536KByte总提交:493            测试通过:175 描述 The digital root of a positive integer is found by summing the digits of…

Digital Roots Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits,…

Digital Roots Problem Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits,…

Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 79339    Accepted Submission(s): 24800 Problem Description The digital root of a positive integer is found by summing the digits of…

Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 67689    Accepted Submission(s): 21144 Problem Description The digital root of a positive integer is found by summing the digits of…

 /*Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 58506    Accepted Submission(s): 18275 Problem Description The digital root of a positive integer is found by summing the digi…

C - Digital Roots Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, thos…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1013 Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 90108    Accepted Submission(s): 28027 Problem Description The digital root of a…

Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 79180    Accepted Submission(s): 24760 Problem Description The digital root of a positive integer is found by summing the digits of…

Digital Roots Time Limit: 1000ms   Memory limit: 65536K 题目描述 The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value co…

先上题目: Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 39890    Accepted Submission(s): 12286 Problem Description The digital root of a positive integer is found by summing the digi…

Digital Roots Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 25766   Accepted: 8621 Description The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that dig…

[九度OJ]题目1124:Digital Roots 解题报告 标签(空格分隔): 九度OJ 原题地址:http://ac.jobdu.com/problem.php?pid=1124 题目描述: The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the d…

2017-09-07 22:02:01 writer:pprp 简单的水题,但是需要对最初的部分进行处理,防止溢出 /* @theme: hdu 1013 Digital roots @writer:pprp @begin:21:52 @end:21:59 @error:虽然是水题,但是还是需要对最初的处理,如果过大超过了int范围,就出错了 @date:2017/9/7 */ #include <bits/stdc++.h> using namespace std; int main() {…

下午做了NYOJ-424Eddy's digital Roots后才正式接触了九余定理,不过这题可不是用的九余定理做的.网上的博客千篇一律,所以本篇就不发篇幅过多介绍九余定理了: 但还是要知道什么是九余定理: 九余数定理 一个数对九取余后的结果称为九余数. 一个数的各位数字之和相加后得到的<10的数字称为这个数的九余数(如果相加结果大于9,则继续各位相加) 简单的说就是:一个整数模9的结果与这个整数的各位数字之和模9的结果相同: 以前做题不知道有这个定理一般暴力就过了,求数位和也不复杂,只不过更…

我在网上看了一些大牛的题解,有些知识点不是太清楚, 因此再次整理了一下. 转载链接: http://blog.csdn.net/iamskying/article/details/4738838 http://www.2cto.com/kf/201405/297531.html 题目描述:求n^n次的digital root(数根),例如root(67)=6+7=root(13)=1+3=4; 一类解法: 求解思路:现在分析一个问题,假设将十位数为a,个位数为b的一个整数表示为ab,则推导得ab…

HDU 1163 题意简单,求n^n的(1)各数位的和,一旦和大于9,和再重复步骤(1),直到和小于10. //方法一:就是求模9的余数嘛! (228) leizh007 2012-03-26 21:03:19 (确实可行) #include<stdio.h> #include<string.h> int main() { int n,i,ans; while(scanf("%d",&n),n) { ans=; ;i<n;i++) { ans=(a…

http://acm.hdu.edu.cn/showproblem.php?pid=1163 九余数定理: 如果一个数的各个数位上的数字之和能被9整除,那么这个数能被9整除:如果一个数各个数位上的数字之和被9除余数是几,那么这个数被9除的余数也一定是几. 证明: 首先10^i ＝99...9(i个9) +1除以9的余数＝1 所以ai*10^i除以9的余数＝ai 用a0~an表示各位数字则数＝(anan-1an-2.a2a1a0), ＝an*10^n+an-1*10^n-1 +an-2 *10^n…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1163 九余数:一个数除于9所得到的余数,即模9得到的值 求九余数: 求出一个数的各位数字之和,如果是两位数以上,则再求各位数字之和,直到只有一位数时结束. 如果求出的和是9,则九余数为0:如果是其他数,则这个数为九余数. 解题思路:快速幂+九余数 先求出九余数,如果九余数为0,则ans=9:如果是其他数,则ans=九余数. #include <iostream> using namespace s…

Background The digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resulting value contains two or more digits, those digits are summed…