杭电acm1013 hdu-acm-1013解题报告

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原题内容：

Digital Roots

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48447    Accepted Submission(s): 15059

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

24390

 

Sample Output

63

 

        本题理解：这一题比较简单，就是输入一个多位数（可能上千位，要用到大数的处理方式）,计算各位的和，如果还是多位数，继续计算；直到是一位数为止。

   代码：

#include<stdio.h>#include<string.h>int z(int a){int sum=0;sum+=a%10;for(;a!=0;){a/=10;sum+=a%10;}return sum;}int main(){int n,sum,i;char c[1010];while(~scanf("%s",c)&&c[0]!='0')//用到大数加减，开始没看清题，wa了； {for(i=0,sum=0;i<strlen(c);i++){sum+=c[i]-'0';}while(sum>9)sum=z(sum);printf("%d\n",sum);}return 0;}