hdu_1013_Digital Roots_201310121652

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 40786    Accepted Submission(s): 12584

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

 

第一次做的：（没考虑大数据的问题）

 

#include <stdio.h>

int f(int t)
{
    int s=0;
    while(t>0)
    {
        s+=t%10;
        t/=10;
    } 
    return s;
}

int main()
{
    int n;
    while(scanf("%d",&n),n)
    {
        int i,j,t,s;
        t=n;
        if(t<10)
        printf("%d\n",t);
        else if(t>=10)
        {
            s=f(t);
            while(s>=10)
            {
                s=f(s);
            }
            printf("%d\n",s);
        }        
    }
    return 0;
}
//wa

第二次做的：

#include <stdio.h>
#include <string.h>

char s[10001];

int f(int t)
{
    int s=0;
    while(t>0)
    {
        s+=t%10;
        t/=10;
    } 
    return s;
}

int main()
{
    while(scanf("%s",s)&&s[0]!='0')
    {
        int i,t,sum=0;
        for(i=0;i<strlen(s);i++)
        sum+=s[i]-'0';
        if(sum<10)
        printf("%d\n",sum);
        else if(sum>=10)
        {
            t=f(sum);
            while(t>=10)
            {
                t=f(t);
            }
            printf("%d\n",t);
        }        
    }
    return 0;
}
//ac

链接（大神做法）：http://www.cppblog.com/ArcTan/articles/167330.html

hdu1013（模拟&数论）

http://acm.hdu.edu.cn/showproblem.php?pid=1013


这个题模拟也可以AC，刚开始我也是模拟AC的。不过看了百度看了大牛的博客，感谢大牛，知道了还有数论这回事。

n=0 1 2 3 4 5 6 7 8 9 10 11 12 13 ......... 100 101 102 103 ....
roots=0 1 2 3 4 5 6 7 8 9 1 2 3 4 .......1 2 3 4....
原来是以1.....9为循环节的。想想也是，每次增加1，那么层层迭代下来，最终当ans<10的时候也是每次增加了1。如此，可以归纳出
roots=(n-1)%9+1

注意输入的数字很大需要字符串读入，求和得n:

#include<stdio.h>
#include<string.h>
#include<math.h>

int main()
{
    int i,n,tmp;
    char a[1003];
    while (scanf("%s",&a)&&a[0]!='0')
    {
        n=0;
        for (i=0;i<strlen(a); i++)
        {
            n+=a[i]-48;
        }
        printf("%d\n",(n-1)%9+1);
    }
    return 0;
}