CF 1064B Equations of Mathematical Magic（思维规律）

Description

Colossal! — exclaimed Hawk-nose. — A programmer! That's exactly what we are looking for.

Arkadi and Boris Strugatsky. Monday starts on Saturday

Reading the book "Equations of Mathematical Magic" Roman Oira-Oira and Cristobal Junta found an interesting equation: $\mathrm{a-(a\oplus x)-x=0}$

for some given $\mathrm{a,\; where\; \oplus \; stands\; for\; a\; bitwise\; exclusive\; or\; (XOR)\; of\; two\; integers\; (this\; operation\; is\; denoted\; as\; ^\; or\; xor\; in\; many\; modern\; programming\; languages).\; Oira-Oira\; quickly\; found\; some\; x}$

, which is the solution of the equation, but Cristobal Junta decided that Oira-Oira's result is not interesting enough, so he asked his colleague how many non-negative solutions of this equation exist. This task turned out to be too difficult for Oira-Oira, so he asks you to help.

Input

Each test contains several possible values of $a$

$\mathrm{and\; your\; task\; is\; to\; find\; the\; number\; of\; equation\text{'}s\; solution\; for\; each\; of\; them.\; The\; first\; line\; contains\; an\; integer\; t\; (1\le t\le 1000}$

) — the number of these values.

The following $t$

$\mathrm{lines\; contain\; the\; values\; of\; parameter\; a,\; each\; value\; is\; an\; integer\; from\; 0\; to\; 230-1}$

inclusive.

Output

For each value of $a$

$\mathrm{print\; exactly\; one\; integer —\; the\; number\; of\; non-negative\; solutions\; of\; the\; equation\; for\; the\; given\; value\; of\; the\; parameter.\; Print\; answers\; in\; the\; same\; order\; as\; values\; of\; a}$

appear in the input.

One can show that the number of solutions is always finite.

Sample Input

Input

3
0
2
1073741823

Output

1
2
1073741824


题目意思：已知a,求解方程$\mathrm{a-(a\oplus x)-x=0，x的可能情况有几种。解题思路：通过移项我们可以得到a\oplus x=a-x，那么我们需要找一下这两个表达式的联系和区别。因为是异或，我们将这两个数放在二进制下比较。}$

   1^1=0 1-1=0
   1^0=1 1-0=1
   0^0=0 0-0=0
   0^1=1 0-1=1//借位

我们可以发现,当a=1时，不管对于位上的x是0还是1，都是成立的，但当a=0时，只有x=0成立。
因而发现a=1时可以有两种选择，那么只需要统计a的二进制中1的个数，根据乘法原则就能求出所有解的个数了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
/*
1^1=0 1-1=0
1^0=1 1-0=1
0^0=0 0-0=0
0^1=1 0-1=1//借位
*/
int main()
{
   int a,b,t,ans;
   scanf("%d",&t);
   while(t--)
   {
     scanf("%d",&a);
     ans=1;
     while(a)
     {
         if(a%2)
         {
             ans=ans*2;
         }
         a/=2;
     }
     printf("%d\n",ans);
   }
   return 0;
}