Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

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Funny Encryption Method

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

---

#include <stdio.h>

int main() {
    int t;
    char s[10];
    scanf("%d", &t);
    while(t--) {
        scanf("%s", s);
        int x1, x2, b1 = 0, b2 = 0;
        sscanf(s, "%d", &x1);
        sscanf(s, "%x", &x2);
        while(x1)
            b1 += x1&1, x1 /= 2;
        while(x2)
            b2 += x2&1, x2 /= 2;
        printf("%d %d\n", b1, b2);
    }
    return 0;
}