hdu1 1001 Mod, Or and Everything 找规律

打表，i从小到大表达二次方，发现输入的$$n<2^i$$时，输出结果为2^(i-1) -1

accode:

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <vector>
#include <algorithm> 
using namespace std;
typedef long long ll;
const ll maxn = 1e12 + 5;
ll a[50];

inline void init()
{
    a[0] = 2;
    ll c = 2, i;
    for(i = 1; c < maxn; i++)
    {
        c *= 2;
        a[i] = c;
    }
}

int main()
{
    init();
    int T;
    ll n;
    scanf("%d", &T);
    cout<<a[1]<<endl;
    while(T--)
    {
        scanf("%lld", &n);
        if(n == 1 || n == 2)
                cout<<0<<endl;
        else
        {
            for(int i = 0; i < 50; i++)
            {
                if(n <= a[i])
                {
                    cout<<a[i-1]-1<<endl;
                    break;
                }
            }
        }        
    }
    
    return 0;
}