SDUT 1048 Digital Roots

最新推荐文章于 2024-05-05 13:24:09 发布

转载最新推荐文章于 2024-05-05 13:24:09 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/yspworld/p/3922370.html

本文介绍了一个简单的算法，用于计算任意正整数的数字根。数字根是指通过不断累加数字直至得到一位数的过程。文章提供了完整的C语言实现代码，并附带了示例输入输出。

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 contains two or more digits, those digits are summed and the process is repeated. This is continued as long as necessary to obtain a single digit.

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.

输入

The input file will contain a list of positive integers, one per line. The end of the input will be indicated by an integer value of zero.

输出

For each integer in the input, output its digital root on a separate line of the output.

示例输入

24
39
0

示例输出

6
3

提示

请注意，输入数据长度不超过20位数字。

题目要求：给你一串数字，这些数字的“和”，如果大于一位数，则再将他们的和数字的每一位再相加，如此下去，直到数字和为一位数字！

算法思想很简单，多用了几个while循环，关键是思路要清晰！

#include <stdio.h>
#include <string.h>

int main()
{
    char s[50];
    int a[50], e;
    int i, j, k;
    int sum;
    while(scanf("%s", s)!=EOF )
    {
        if(s[0]=='0')
        break;
        int len=strlen(s);
        e=0;
        sum=0;
        for(i=0; i<len; i++)
        {
            a[e++]=s[i]-48 ;
            sum+=a[e-1];
        }
        if(sum<10)
            printf("%d\n", sum );
        else
        {
            int cnt=sum;
            sum=0;
            while(cnt>10)
            {
                while(cnt!=0)
                {
                    sum=sum + (cnt%10);
                    cnt=cnt/10;
                }
                if(sum<10)
                {
                    printf("%d\n", sum );
                    break;
                }
                else
                {
                    cnt=sum;
                    sum=0;
                }
            }
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/yspworld/p/3922370.html