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  • ACM HDU 1013 Digital Roots

    Digital Roots

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 20786    Accepted Submission(s): 6161


    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
     

    Sample Output
    6 3
     

    Source
     
     
     
    这道题很简单,但是有陷阱,就是 数据很大,不能用int存 ,要用字符串读取!
    //要用高精度 
    #include<stdio.h>
    #include
    <string.h>
    int main()
    {
    int i,n,d;
    char ch[1000];//这个一定要开得够大
    memset(ch,0,sizeof(ch));
    while(scanf("%s",&ch)&&strcmp(ch,"0")!=0)
    {
    int len=strlen(ch);
    n
    =0;
    for(i=0;i<len;i++)
    n
    +=ch[i]-'0';

    while(n/10!=0)
    {
    d
    =0;
    while(n)
    {
    d
    +=n%10;
    n
    /=10;
    }
    n
    =d;
    }
    printf(
    "%d\n",n);
    }
    return 0;
    }

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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2118720.html
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