【CUGBACM15级BC第22场 A】hdu 5142 NPY and FFT_【cugbacm15级bc第22场 c】hdu 5144 npy and shot-优快云博客

本文介绍了一个有趣的二进制操作问题：如何将一个整数转换为二进制形式，然后翻转该二进制数并去除前导零，最后将其重新解释为十进制数。通过具体示例和C++实现代码详细展示了这一过程。

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NPY and FFT

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1064 Accepted Submission(s): 699

Problem Description

A boy named NPY is learning FFT algorithm now.In that algorithm,he needs to do an operation called "reverse".
For example,if the given number is 10.Its binary representaion is 1010.After reversing,the binary number will be 0101.And then we should ignore the leading zero.Then the number we get will be 5,whose binary representaion is 101.
NPY is very interested in this operation.For every given number,he want to know what number he will get after reversing.Can you help him?

Input

The first line contains a integer T — the number of queries (

1≤T≤100 ).
The next T lines,each contains a integer

X(0≤X≤231−1),the given number.

Output

For each query,print the reversed number in a separate line.

Sample Input

Sample Output

3
1
1

题意：把n转换为二进制再翻转，去掉前导零后的数是多少

思路：用数组保存二进制，处理前导零即可

#include <iostream>
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <algorithm>
#include <functional>
#include <ctime>
#define PI acos(-1)
#define eps 1e-8
#define inf 0x3f3f3f3f
#define debug(x) cout<<"---"<<x<<"---"<<endl
typedef long long ll;
using namespace std;

long long quickpow(long long a, long long b)
{
    if (b < 0)
    {
        return 0;
    }
    long long ret = 1;
    for (; b; b >>= 1, a = (a * a))
        if (b & 1)
        {
            ret = (ret * a);
        }
    return ret;
}
int a[55];
int main()
{
    std::ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--)
    {
        memset(a, 0, sizeof(a));
        int x;
        cin >> x;
        int cnt = 0, flag;
        while (x)
        {
            a[++cnt] = x % 2;
            x /= 2;
        }
        for (int i = 1; i <= cnt; i++)
        {
            if (a[i] == 1)
            {
                flag = i;
                break;
            }
        }
        ll ans = 0;
        for (int i = flag; i <= cnt; i++)
        {
            ans += quickpow(2, cnt - i) * a[i];
        }
        cout << ans << endl;
    }
    return 0;
}