UVA10469 To Carry or not to Carry【数字逻辑运算】

6+9=15 seems okay. But how come 4+6=2?
  You see, Mofiz had worked hard throughout his digital logic course, but when he was asked to implement a 32 bit adder for the laboratory exam, he did some mistake in the design part. After tracing the design for half an hour, he found his flaw!! He was doing bitwise addition but his carry bit always had zero output. Thus,


  4 = 00000000 00000000 00000000 00000100
+6= 00000000 00000000 00000000 00000110
----------------------------------------
 2 = 00000000 00000000 00000000 00000010
  Its a good thing that he finally found his mistake, but it was too late. Considering his effort throughout the course, the instructor gave him one more chance. Mofiz has to write an efficient program that would take 2 unsigned 32 bit decimal numbers as input, and produce an unsigned 32 bit decimal number as the output adding in the same was as his circuit does.
Input
In each line of input there will be a pair of integer separated by a single space. Input ends at EOF.
Output
For each line of input, output one line — the value after adding the two numbers in the ‘Mofiz way’.
Sample Input
4 6
6 9
Sample Output
2
15

问题链接：UVA10469 To Carry or not to Carry
问题简述：（略）

问题分析：

  仔细看一下原题的算式可以知道如下：
    0+0=0
    0+1=1
    1+0=1
    1+1=0
很容易看出来是异或运算。

 


  其实，看电路可以看出来是一个加法器电路，其和计算就是异或。
程序说明：（略）
题记：（略）
参考链接：（略）

AC的C++语言程序如下：/* UVA10469 To Carry or not to Carry */

#include <bits/stdc++.h>

using namespace std;

int main()
{
int a, b;

while(~scanf("%d%d", &a, &b))
printf("%d\n", a ^ b);

return 0;
}

/*
0+0=0
0+1=1
1+0=1
1+1=0
*/