UVa 10019 Funny Encryption Method (water ver.)

10019 - Funny Encryption Method

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=960

The Problem

History :

A student from ITESM Campus Monterrey plays with a new encryption method for numbers. These method consist of the following steps:

Steps : Example

1) Read the number N to encrypt M = 265

2) Interpret N as a decimal number X1= 265 (decimal)

3) Convert the decimal interpretation of N to its binary representation X1= 100001001 (binary)

4) Let b1 be equal to the number of 1’s in this binary representation B1= 3

5) Interpret N as a Hexadecimal number X2 = 265 (hexadecimal)

6) Convert the hexadecimal interpretation of N to its binary representation X2 = 1001100101

7) Let b2 be equal to the number of 1’s in the last binary representation B2 = 5

8) The encryption is the result of M xor (b1*b2) M xor (3*5) = 262
This student failed Computational Organization, that’s why this student asked the judges of ITESM Campus Monterrey internal ACM programming Contest to ask for the numbers of 1’s bits of this two representations so that
he can continue playing.
Task :

You have to write a program that read a Number and give as output the number b1 and b2

The Input

The first line will contain a number N which is the number of cases that you have to process. Each of the following N Lines ( 0<N<=1000) will contain the number M (0<M<=9999, in decimal representation) which is
the number the student wants to encrypt.

The Output

You will have to output N lines, each containing the number b1 and b2 in that order, separated by one space corresponding to that lines number to crypt

Sample Input

3

265

111

1234

Sample Output

3 5

6 3

5 5

完整代码：

/*0.016s*/

#include<cstdio>

inline int bitnum(int num)
{
	int count = 0;
	while (num)
	{
		if (num & 1) ++count;
		num >>= 1;
	}
	return count;
}

int main()
{
	int t, m, b1, b2;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &m);
		b1 = bitnum(m), b2 = 0;
		while (m)
		{
			b2 += bitnum(m % 10);
			m /= 10;
		}
		printf("%d %d\n", b1, b2);
	}
	return 0;
}