zoukankan      html  css  js  c++  java
  • 杭电oj Problem-1013 Digital Roots

    Digital Roots

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 65477    Accepted Submission(s): 20417


    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
     
    #include <stdio.h>
    #include <string.h>
    int main()
    {
        char a[10000005];   //数组要开大
        int len, len1, k, i;
        while (scanf("%s", a) != EOF)
        {
            len = strlen(a);
            if (a[0] == '0' && len == 1)  break;
            int temp;
            int sum = 0;
            for(i = 0;i < len; i++)    sum += a[i] - '0';
            while (sum > 9) {
                k = 1;
                for (len1 = 1; ; len1++){
                    k *= 10;
                    if(k > sum)  break;
                }
                temp = 0;
                for(i = 0;i < len1; i++){
                    temp += sum % 10;
                    sum /= 10;
                }
                sum = temp;
            }
            printf("%d
    ", sum);
        }
        return 0;
    }
    
    

  • 相关阅读:
    剑指offer:合并两个排序的链表
    剑指offer:调整数组顺序使奇数位于偶数前面
    剑指offer:链表中倒数第K个结点
    牛客网在线编程:末尾0的个数
    剑指offer7:数值的整数次方
    牛客网在线编程:计算糖果
    牛客网在线编程:求数列的和
    牛客网在线编程:公共字符
    剑指offer7:斐波那契数列
    Qt入门之常用qt控件认知之Button系列
  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770970.html
Copyright © 2011-2022 走看看