poj1519 數論簡單題

Digital Roots

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 26535		Accepted: 8874

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

這道題要求求一個數的“樹根”。先求一個數每位數字之和，如果和不爲一個個位數，則將該數再求位數和，直到位數和爲個位數，這個個位數就是該數的“樹根”。

開始做的時候用的是遞歸寫的，結果超時了。正確解這題的做法是將該數字每位數字之和再mod 9。這就行了。

樹根=位數和（mod 9）

代碼：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
char str[100];
int main()
{
    
    while(scanf("%s",str))
    {
        if(str[0]=='0')
            break;
        int sum=0;
        int len=strlen(str);
        for(int i=0;i<len;i++)
        {
            sum+=str[i]-'0';
        }
        sum=sum%9;
        if(!sum)
            sum=9;
        printf("%d\n",sum);
    }
    return 0;
}