Eddy's digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5928    Accepted Submission(s): 3270 Problem Description The digital root of a positive integer is found by summing the digit…

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…

Digital Roots Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 64420    Accepted Submission(s): 20053 Problem Description The digital root of a positive integer is found by summing the digits of…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1163 思路: 九余数定理:一个数对九取余的结果叫做九余数, 一个数的各个位数相加的得到的小于10的数也叫九余数 #include<iostream> #include<cstdio> #include<cstring> using namespace std; int main(void) { int n,tp,i; while(~scanf("%d",…

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…

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): 4632    Accepted Submission(s): 2578 Problem Description The digital root of a positive integer is found by summing the digi…

题目传送门:HDU1013 九余数定理 //题目描述:给定一个数,要求你求出它的每位上的数字之和,并且直到每位上的数字之和为个位时候输出它 //输入:一个整数 //输出:题目描述的结果 //算法分析: 此题要用到9余数定理. 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 ) #include <iostream> #include <cstdlib> #inclu…

1.HDU1013求一个positive integer的digital root,即不停的求数位和,直到数位和为一位数即为数根. 一开始,以为integer嘛,指整型就行吧= =(too young),后来大数自然用字符串解决,然后get到一个新数论点九余数定理: https://en.wikipedia.org/wiki/Digital_root 即:一个数的数根等于它模 9 的余数.(=>几个数之积的九余数=每个数的九余数之积的九余数.) 2.HDU1163,2035求n^n的数根,即九余…