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

 

Source

Greater New York 2000

 

我还是太年轻了 

我天真的认为 n一定在int范围内 

然后就WA啊WA

 

#include<cstdio>
#include<iostream>

using namespace std;

int n;

inline int find(int x) {
    int ans=0;
    while(x) {
        ans+=x%10;
        x/=10;
    }
    return ans;
}

int main() {
    while(scanf("%d",&n)&&n) {
        do {
            n=find(n);
        }while(n>=10);
        printf("%d
",n);
    }
    return 0;
}

/*
    这是九余数定理 
    什么鬼 好像挺有道理 
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#define MAXN 100010

using namespace std;

char s[MAXN];

int main() {
    while(~scanf("%s",s)) {
        int len=strlen(s),ans=0;
        if(len==1&&s[0]=='0') break;
        for(int i=0;i<len;i++) ans+=s[i]-48;
        if(ans%9==0) printf("9
");
        else printf("%d
",ans%9);
    }
    return 0;
}


/*
    非九余数定理 
        处理一下字符串就好了
*/
#include<cstdio>
#include<cstring> 
#include<iostream>
#define MAXN 100010

using namespace std;

char s[MAXN];

inline int find(int x) {
    int ans=0;
    while(x) {
        ans+=x%10;
        x/=10;
    }
    return ans;
}

int main() {
    while(~scanf("%s",s)) {
        int len=strlen(s),ans=0;
        if(len==1&&s[0]=='0') break;
        for(int i=0;i<len;i++) ans+=s[i]-48;
        if(ans<=9) printf("%d
",ans);
        else {
            do {
                ans=find(ans);
            }while(ans>=10);
            printf("%d
",ans);
        }
    }
    return 0;
}