算法小题目小结。。。

一、Digital Roots

HDOJ地址： http://acm.hdu.edu.cn/showproblem.php?pid=1013

不过数学规律是大神！

代码如下：

/***** Digital Roots********/

/******** written by C_Shit_Hu ************/

////////////////简单大数///////////////

/****************************************************************************/
/* 
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(the length of each integer will not exceed 1000),
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.
*/
/****************************************************************************/


// 数学规律是大神！！！
#include<stdio.h>
#include<string.h>
char s[1010];
int main()
{
    int sum,i,len,x;
    while(gets(s)&&s[0]!='0')
    {
		sum=0;
		len=strlen(s);
		for(i=0;i<len;i++)
			sum+=s[i]-'0';
		x=sum%9;
		if(x==0) x+=9;
		printf("%d
",x);
    }
    return 0;
}


/******************************************************/
/********************  心得体会  **********************/
/*
说明1.各位数字相加知道一位数字x，这个x等于原来的数除以9的余数（原数为9的倍数除外）
    2.若原数为9的倍数，最后结果x+9就行了
	3.一个数字除以9的余数等于该数各数字之和除以9的余数。（这点为大数运算提供了极大的方便！）

  水水更健康！！！
*/
/******************************************************/

二、Hailstone HOTPO

地址： http://acm.hdu.edu.cn/showproblem.php?pid=4484

简单的大叔题目。类似于3n+1de题目。。。。

代码如下：

/***** Hailstone HOTPO ********/

/******** written by C_Shit_Hu ************/

////////////////简单的数学题目///////////////

/****************************************************************************/
/* 
类似于3n+1的题目
*/
/****************************************************************************/


// 
#include <stdio.h>

_int64 MAX( _int64 num )
{
	_int64 max =num ;
	while(num != 1)
	{
		if (num%2 == 0)
		{
			num /=2 ;
		} 
		else
		{
			num = num * 3 +1 ;
			if (max < num)
			{
				max = num ;
			}
		}
	}
	return max ;
}


int main()
{
	_int64 P, order, num, i ;
	_int64 max ;
	scanf("%I64d", &P) ;
	for (i=0; i<P; i++)
	{
		scanf("%I64d%I64d", &order, &num) ;
		max = MAX( num ) ;
		printf("%I64d %I64d
", order, max) ;
	}
	return 0 ;
}


/******************************************************/
/********************  心得体会  **********************/
/*
果然是水题。。。
还做的那么慢。。

  水水更健康！！！
*/
/******************************************************/