本文主要是介绍CF1064B Equations of Mathematical Magic，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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: a−(a⊕x)−x=0a−(a⊕x)−x=0for some given aa, where ⊕⊕ 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 xx, 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 aa and your task is to find the number of equation's solution for each of them. The first line contains an integer tt (1≤t≤10001≤t≤1000) — the number of these values.
The following tt lines contain the values of parameter aa, each value is an integer from 00 to 230−1230−1 inclusive.
Output
For each value of aa 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 aa appear in the input.
One can show that the number of solutions is always finite.
Example
Input
3 0 2 1073741823
Output
1 2 1073741824
Note
Let's define the bitwise exclusive OR (XOR) operation. Given two integers xx and yy, consider their binary representations (possibly with leading zeroes): xk…x2x1x0xk…x2x1x0 and yk…y2y1y0yk…y2y1y0. Here, xixi is the ii-th bit of the number xx and yiyi is the ii-th bit of the number yy. Let r=x⊕yr=x⊕y be the result of the XOR operation of xx and yy. Then rr is defined as rk…r2r1r0rk…r2r1r0 where:
ri={1, if xi≠yi0, if xi=yiri={1, if xi≠yi0, if xi=yi
For the first value of the parameter, only x=0x=0 is a solution of the equation.
For the second value of the parameter, solutions are x=0x=0 and x=2x=2.
题意：满足a-x=a^x等式x的个数（x非负）
分析：明天补上
代码：
#include<iostream>
#include<cmath>
using namespace std;
int main()
{long long int n,temp,ans;cin>>n;for(int i=1;i<=n;i++){ans=1;cin>>temp;while(temp){if(temp%2){ans*=2;temp/=2;}else{temp/=2;}}cout<<ans<<endl;}return 0;
}
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