B. Asia Ltd. region (hex conversion)

Single Point Test time:  1.0 seconds

Memory Limit:  512 MB

Binary data is a number with 0 and 1, two digital represented. Its base 2, carry rule is "every two into a" borrow rule is "borrow one when two", a master of philosophy in Germany mathematical 18th century Levin Bu Nizi found.
hexadecimal (hex or subscript abbreviated as 16) in a mathematics every 16 into a binary system of general use numbers 0 to 9 and the letters a to F (or a ~ f) represents wherein: a ~ f represents 10 to 15, the hexadecimal numbers called.
Please binary number given to hexadecimal numbers, letters in lower case.

Entry

The first line of a positive integer T, T representatives of groups of test data (1≤T≤10)..
Next T rows each contain only input string  0 and  1 a string (no leading 0), the string length: 1 ≤length≤106.

Export

For each test case, the hexadecimal number after conversion and output on a separate line.

Sample

input

2
1
10

output

1
2

按四位二进制合成一位十六进制来考虑

 

#include<bits/stdc++.h>
#include<algorithm>

using namespace std;
map<int,char> mmp;
char s[1000005];
int a[1000000];

int main()
{
	int t;
	scanf("%d",&t);
	mmp[10] = 'a';
	mmp[11] = 'b';
	mmp[12] = 'c';
	mmp[13] = 'd';
	mmp[14] = 'e';
	mmp[15] = 'f';
	while (t--)
	{
		int x = 0;
		scanf("%s",s+1);
		int len = strlen(s+1);
		int k = 0,m = len / 4;
		int sum = 0;
		if (len<=4)//当输入不大于四位是可以直接输出
		{
			for (int i=len;i>=1;i--)
			{
				int p = s[i] - '0';
				for (int j=0;j<k;j++)//k用来表示每一位x2的个数（倒着计算）
					p *= 2;
				sum += p;;
				k++;
				if (i==1)
				{
					a[x++] = sum;
					break;
				}
			}
			if (sum<=9)
				printf("%d\n",sum);
			else
				printf("%c\n",mmp[sum]);
			continue;
		}
		k = 0,sum = 0;
		for (int i=len;i>len-m*4;i--)
		{
			int p = s[i] - '0';
			for (int j=0;j<k;j++)
				p *= 2;
			sum += p;;
			k++;
			if (k==4)
			{
				k = 0;
				a[x++] = sum;
				sum = 0;
			}
		}
		k = 0,sum = 0;
		for (int i=len-m*4;i>=1;i--)
		{
			int p = s[i] - '0';
			for (int j=0;j<k;j++)
				p *= 2;
			sum += p;;
			k++;
			if (i==1)
			{
				a[x++] = sum;
				sum = 0;
				break;
			}
		}
		for (int i=x-1;i>=0;i--)
		{
			if (a[i]<=9)
				printf("%d",a[i]);
			else
				printf("%c",mmp[a[i]]);
		}
		puts("");
	}
	return 0;
}