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 and a new encryption method for numbers. These are 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 is 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 is 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 is student failed computational organization, that's why this student asked the judges of ITESM campus Monterrey Interna L ACM Programming Contest to ask for the numbers of 1 ' s bits of this two representations so and he can continue playing.

Task:
You are have to write a program that read a number and give as output the number B1 and B2

The Input

The the A number N which is the number of cases that you have to process. Each of the following N Lines (0<n<=1000) would contain the number M (0<m<=9999, in decimal representation) W Hich is the number of the student wants to encrypt.

The Output

This article URL address: http://www.bianceng.cn/Programming/sjjg/201410/45353.htm

You'll have to output N lines, containing of the number B1 and B2 in, separated by one space corresponding To, lines number to crypt

Sample Input

3
265
111
1234

Sample Output

3 5
6 3

5 5

Complete code:

/*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);  
            M/=;  
        printf ("%d%d\n", B1, B2);  
    }  
    return 0;  
}