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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): 31003 Accepted Submission(s): 9523


    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
     
     1 #include<stdio.h>
     2 #include<string.h>
     3 char digit[1000];
     4 int main()
     5 {
     6     int i, j, len, sum, n;
     7     memset(digit, 0, sizeof(digit));
     8     while(scanf("%s", digit) && strcmp(digit, "0") != 0)
     9     {
    10         n = sum = 0;
    11         len = strlen(digit);
    12         for(i=0; i < len; i++)
    13         {
    14             n = n + (digit[i] - '0');
    15         }
    16         while(1)
    17         {
    18             sum += n%10;
    19             n /= 10;
    20             if(n == 0)
    21                 if(sum >= 10) 
    22                 {
    23                     n = sum;
    24                     sum = 0;
    25                 }
    26                 else break;
    27         }
    28         printf("%d\n", sum);
    29         memset(digit, 0, sizeof(digit));
    30     }
    31     return 0;
    32 }

    解题报告:

    没事少用fgets,要不水题在菜鸟的眼里都是炙手可热的东西。

    物役记

    更多内容请关注个人微信公众号 物役记 (微信号:materialchains)

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  • 原文地址:https://www.cnblogs.com/liaoguifa/p/2700543.html
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