zoukankan      html  css  js  c++  java
  • hdu 1013 Digital Roots

    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

    原来这个题是大数问题,唉!提交了两次,WA了才发现这个问题

    #include<map>
    #include<set>
    #include<queue>
    #include<cmath>
    #include<vector>
    #include<cstdio>
    #include<string>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #define  inf 0x0f0f0f0f
    
    using namespace std;
    
    const double pi=acos(-1.0);
    const double eps=1e-8;
    typedef pair<int,int>pii;
    
    
    int main()
    {
        //freopen("in.txt","r",stdin);
    
        char str[10000];
        scanf("%s",str);
        while (!(strlen(str)==1 && str[0]=='0'))
        {
            int len=strlen(str);
            long long sum=0;
            for (int i=0;i<len;i++) sum+=str[i]-'0';
            while (sum/10)
            {
                long long n=sum;
                sum=0;
                while (n)
                {
                    sum+=n%10;
                    n=n/10;
                }
            }
            printf("%lld
    ",sum);
            scanf("%s",str);
        }
        //fclose(stdin);
        return 0;
    }
    至少做到我努力了
  • 相关阅读:
    babel-loader的原理
    js弱类型转换的知识点
    个人说明
    嗯三月的小计划
    bzoj2326 [HNOI2011]数学作业
    bzoj2654 tree
    bzoj1018[SHOI2008]堵塞的交通traffic
    bzoj1937[Shoi2004]Mst 最小生成树
    一般图最大匹配-带花树-带注释模板
    FFT快速傅里叶变换-递归版-带注释模板
  • 原文地址:https://www.cnblogs.com/chensunrise/p/3690322.html
Copyright © 2011-2022 走看看