[Exercises] algorithm notes issue C: hex conversion

Ordinary decimal to binary approach is to take the remainder divisor method.

0 digit decimal number, later found Now type long long int range of about 10 18 ^, 10 ^ 30 indicates insufficient. So the idea this problem is to use an array to store the value, and then use an array of ways modulo divisor simulation method:

① array The last number is the number of decimal mantissa, mantissa value of the modulo 2

The decimal representation of the array ② constantly divided by two.

Examples: b [] to store a decimal number, such as 13, so b [0-1] = {1,3}, num [] array to store the mantissa 2% value, i.e. the final binary storage reverse order

a.13, mantissa 3,3% 2 = 1, num [0] = 1;

b.12 tell me is 1,1 / 2 = 0, since the first is an odd number, an odd number not divisible by 2, the lower 3 + 10 = 13, so the low number by, (3 + 10) / 2 = 6,

  Whereby B [] array into b [] = {0,6}, recycled ab, sequentially into an array of values:
 {0,6} -> {0,3} -> {0, 1} -> { 0,0}, during num [] = {1,0,1,1}

So 13 = (1101) 2;

The code is given below:

#include <stdio.h>
#include <string.h>

int main()
{
	
	char snum[32];
	int b[32]; 
	
	while(scanf("%s", snum) != EOF)
	{
		int i=0,len = strlen(snum);
		//字符串转为整数,存到数组b[]中 
		for(i=0; i<len; i++)
			 b[i] = snum[i] -'0'; 

		//对数组b[]模拟除商取余
		i=0;
		char num[200];
		int k=0, cf, j,temp;  //cf是进位 
		while(i<len)
		{
			num[k++] = (b[len-1] % 2) + '0';  //尾数取余，并存到二进制表示数组中

            //对整个剩余的值除以2,从第i个位置到结束  
			cf=0; //每一轮cf都要清0 
			for(j=i; j<len; j++)
			{
				temp = b[j]; 
				b[j] = (b[j] + cf)/2; //b[j]+cf进位10 
				if(temp % 2 == 1)
					cf = 10;
				else
					cf = 0;		 
			}
			if(b[i] == 0) //高位如果变为0，处理下一个位置 
				i++;
			 
		}
		
		//输出
		for(j=k-1; j>=0;j--)
			printf("%c",num[j]);
			 
		printf("\n");
	}
	
	
	return 0;
}