Digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
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2023-05-05 18:26:12
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1 #include 2 #include 3 #include 4 #include 5 #include 6 using namesp
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2022-05-25 10:39:47
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Problem DescriptioThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then ...
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2014-03-29 14:38:00
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DescriptionThe 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 valu
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2023-04-14 00:28:09
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ots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) T
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2017-07-07 15:42:00
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Digital RootsTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 29353Accepted Submission(s): 8974Problem DescriptionThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then tha
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2021-07-29 16:26:04
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题目大意：将给出的一个大数各位数字相加，得到一个新的数，再将这个数字各位数字相加，依次计算，直到变成一个个位数，这个数就被称为这个大数的Digital Roots。现在要求得到这个数。单词积累：digit 位，数字题解：其实这道题目根本就不需要高精度计算，一开始，直接将各位数字相加，和就在int的范围内了，然后就可以直接算了。#include #include int main(){ char s[1000]; int n; while(scanf("%s",s),s[0]!='0') { n=0; for(int i=0; i9...
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2014-01-17 16:28:00
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传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1013水题我的方法是当做字符串进行处理，算完和之后，再用一次sprintf，将int数值转换为字符串，在进行求和操作，直到满足条件。。。。。#include<stdio.h>#include<string.h>int main() { char s[10001]; memset(s, '
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2021-06-29 14:43:02
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BackgroundThe 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
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2017-03-08 18:01:00
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Digital RootsProblem DescriptionThe 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 and the process is repeated.
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2014-03-24 18:08:00
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传送门：http://acm.hdu.edu.cn/showproblem.php?pid=1013水题我的方法是当做字符串进行处理，算完和之后，再用一次sprintf，将int数值转换为字符串，在进行求和操作，直到满足条件。。。。。#include<stdio.h>#include<string.h>int main() { char s[10001]; memset(s, '
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2022-02-15 11:15:12
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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
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2022-05-14 11:47:34
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Digital Roots Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)Total Submission(s) : 19 Accepted Submission(s) : 7Font: Times New Roman | Verdana | Georgia Fon
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2022-12-02 00:28:44
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题目链接：hdu 1013 Digital Roots 题目大意:给出一个数num，计算num每一个位数上的值之和，的到新的num，如果num>= 10, 重复操作，直到为个数。 解题思路:第一次输入的时候位数很多，得用字符串读入。 #include #include #include const int N = 1
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2013-11-01 18:39:00
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Digital Roots Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 68194 Accepted Submission(s): 21324
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2016-05-21 09:55:00
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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 result...
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2023-05-09 10:01:00
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