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  • Digital Roots 分类: HDU 2015-06-19 22:56 13人阅读 评论(0) 收藏

    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 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.

    Input
    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.

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

    Sample Input

    24 39 0
    在自己的OJ字符串开到20就过了,在这开到110还RE,醉啊

    #include <iostream>
    #include <iomanip>
    #include <cstdio>
    #include <cstring>
    using namespace std;
    int main()
    {
        char s[1110];
        int sum;
        while(cin>>s)
        {
            if(strcmp(s,"0")==0)
                break;
            int len=strlen(s);
            sum=0;
            for(int i=len-1;i>=0;i--)
            {
                sum+=(s[i]-'0');
                if(sum>9)
                {
                    sum=sum%10+sum/10;
                }
            }
            cout<<sum<<endl;
        }
        return 0;
    }
    

    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4722005.html
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