Digital Roots
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Digital Roots
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 65092 Accepted Submission(s): 20292
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.
本题我刚开始没有考虑输入的整数无穷大情况,就固定了输入数据类型,后来改为string,但是求得和还是可能无穷大,这一点没考虑但是程序居然通过了,汗
#include<iostream>
#include<string>
using namespace std;
int getRootInt(int value)
{
int sum=0;
int tempValue=value;
while(tempValue!=0)
{
sum+=tempValue%10;
tempValue=tempValue/10;
}
if(sum>9)
{
sum=getRootInt(sum);
}
return sum;
}
int getRoot(string value)
{
int sum=0;
for(int i=0;i<value.length();i++)
{
sum+=value[i]-‘0‘;
}
return getRootInt(sum);
}
int main()
{
string value="";//不能定义固定类型可能无穷大
while(cin>>value)
{
if(value=="0")break;
int sum=0;
sum=getRoot(value);
cout<<sum<<endl;
}
return 0;
}
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