process clu auto white art cst int posit val

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

一個數。各個位數相加得到的數假設小於10就輸出，否則就繼續把得到的數各個位數相加。一看我就模擬做的。模擬的時候要註意。輸入的數字可能非常大。所以用int是不能夠的，要用字符串處理。模擬做法代碼例如以下：

#include<cstdio>
#include<cstring>
void zuo(int x){
    int sum=0;
    while(x){
        sum+=(x%10);
        x/=10;
    }
    if(sum<10) printf("%d\n",sum);
    else zuo(sum);
}
int main(){
    char s[1010];
    while(scanf("%s",s)&&s[0]!=‘0‘){
        int x=0;
        for(int i=0;i<strlen(s);i++)
            x+=(s[i]-‘0‘);
        zuo(x);
    }
    return 0;
}

另一種解法。數論的知識。

數字本身： 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 12 22 23 24 25 26 27 28 29 30············

各個位數和： 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3·············

你會發現。每9個是一個循環。所以僅僅要對9取余就ok了。代碼例如以下：

#include<cstdio>
#include<cstring>
int main(){
    char s[1010];
    while(scanf("%s",s)&&s[0]!=‘0‘){
        int x=0;
        for(int i=0;i<strlen(s);i++)
            x+=(s[i]-‘0‘);
        x=x%9;
        if(x==0) printf("9\n");
        else printf("%d\n",x);
    }
    return 0;
}

HDU 1013.Digital Roots【模擬或數論】【8月16】