Digital Roots(杭电1013)(字符串处理)(大数)
发布时间:2020-12-14 02:55:24 所属栏目:大数据 来源:网络整理
导读:Digital Roots Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 50923????Accepted Submission(s): 15890 Problem Description The digital root of a positive integer is found by summing th
Digital RootsTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 50923????Accepted Submission(s): 15890
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.
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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.
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Output
For each integer in the input,output its digital root on a separate line of the output.
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Sample Input
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Sample Output
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Source
Greater New York 2000
#include<stdio.h> #include<string.h> int fenjie(int n) { int ans=0; while(n>9) { ans+=n%10; n/=10; } ans+=n; return ans; } int main() { char s[1010]; int i,len,sum; while(gets(s)&&s[0]!='0') { sum=0; len=strlen(s); for(i=0;i<len;i++) { sum+=(s[i]-'0'); } while(sum>9) { sum=fenjie(sum); } printf("%dn",sum); } return 0; } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |