hdu 1013 Digital Roots

最新推荐文章于 2021-05-19 20:19:13 发布

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最新推荐文章于 2021-05-19 20:19:13 发布

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分类专栏： # ACM-算法设计

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本文介绍了一种计算数字根的有效方法，通过将输入整数转换为字符串，并对其每一位进行求和来计算数字根，当和大于9时，进一步优化求和过程以提高效率。

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题目链接：点击打开链接

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

Sample Output

6
3

ps:此题用数组暴力定会超时，所以想到字符串，但是只是用字符串第一遍加和还是超时，所以要优化，再每次和超过9时都进行拆分加和从而减少时间

<span style="font-size:18px;">#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char str[10000000];
int main()
{
    int n;

    while(scanf("%s",str))
    {
        if(strcmp(str,"0")==0)break;
        n=0;
        int len=strlen(str);
        for(int i=0; i<len; i++ )
        {
            n=n+(int)str[i]-'0';
            if(n>9)
            {
                n=n%10+n/10;
            }
        }
        printf("%d\n",n);
    }
    return 0;
}
</span>