hdu1013(基础)
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Digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 65684 Accepted Submission(s): 20476
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.
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
刚开始以为数输入的就是一个整型数,没想到提交之后就wa,后来用数组试了一下过了。
AC代码:
//hdu(1013)//题目大意:就一个数的数根,当一个数各位数字之和sum是一个个位数,那么其就是该数的数根。//否则,继续求sum的各位数之和,直到得到一个个位数,该个位数就是这个数的数根。 #include<stdio.h>#include<string.h>char a[1010];int main(){int n,sum,i,j;while(scanf("%s",a)){sum=n=0;int len=strlen(a);if(a[0]=='0') break;for(i=0;i<len;i++) n+=a[i]-'0';sum=n;if(n>=10){do{sum=0;while(n){sum+=(n%10);n/=10;}n=sum;}while(sum>=10);}printf("%d\n",sum);memset(a,0,sizeof(a));}return 0;}
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