zoukankan      html  css  js  c++  java
  • CodeForces

    Can the greatest common divisor and bitwise operations have anything in common? It is time to answer this question.

    Suppose you are given a positive integer aa. You want to choose some integer bb from 11to a−1a−1 inclusive in such a way that the greatest common divisor (GCD) of integers a⊕ba⊕b and a&ba&b is as large as possible. In other words, you'd like to compute the following function:

    f(a)=max0<b<agcd(a⊕b,a&b).f(a)=max0<b<agcd(a⊕b,a&b).

    Here ⊕⊕ denotes the bitwise XOR operation, and && denotes the bitwise AND operation.

    The greatest common divisor of two integers xx and yy is the largest integer gg such that both xx and yy are divided by gg without remainder.

    You are given qq integers a1,a2,…,aqa1,a2,…,aq. For each of these integers compute the largest possible value of the greatest common divisor (when bb is chosen optimally).

    Input

    The first line contains an integer qq (1≤q≤1031≤q≤103) — the number of integers you need to compute the answer for.

    After that qq integers are given, one per line: a1,a2,…,aqa1,a2,…,aq (2≤ai≤225−12≤ai≤225−1) — the integers you need to compute the answer for.

    Output

    For each integer, print the answer in the same order as the integers are given in input.

    Example

    Input

    3
    2
    3
    5
    

    Output

    3
    1
    7
    

    Note

    For the first integer the optimal choice is b=1, then a⊕b=3,a&b=0, and the greatest common divisor of 33 and 00 is 33.

    For the second integer one optimal choice isb=2, then a⊕b=1, a&b=2, and the greatest common divisor of 1and 2 is 1.

    For the third integer the optimal choice is b=2, then a⊕b=7, a&b=0a&b=0, and the greatest common divisor of 7 and 0 is 7.

    代码:

    #include<cstdio>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<set>
    #include<map>
    
    using namespace std;
    
    
    
    int main()
    {
    	int n;
    	cin>>n;
    	long long x;
    	for(int t=0;t<n;t++)
    	{
    		scanf("%lld",&x);
    		long long sum=1;
    		if(x==3)
    		{
    			printf("1
    ");
    		}
    		else if(x==7)
    		{
    			printf("1
    ");
    		}
    		else if(x==15)
    		{
    			printf("5
    ");
    		}
    		else if(x==31)
    		{
    			printf("1
    ");
    		}
    		else if(x==63)
    		{
    			printf("21
    ");
    		}
    		else if(x==127)
    		{
    			printf("1
    ");
    		}
    		else if(x==255)
    		{
    			printf("85
    ");
    		}
    		else if(x==511)
    		{
    			printf("73
    ");
    		}
    		else if(x==1023)
    		{
    			printf("341
    ");
    		}
    		else if(x==2047)
    		{
    			printf("89
    ");
    		}
    		else if(x==4095)
    		{
    			printf("1365
    ");
    		}
    		else if(x==8191)
    		{
    			printf("1
    ");
    		}
    		else if(x==16383)
    		{
    			printf("5461
    ");
    		}
    		else if(x==32767)
    		{
    			printf("4681
    ");
    		}
    		else if(x==65535)
    		{
    			printf("21845
    ");
    		}
    		else if(x==131071)
    		{
    			printf("1
    ");
    		}
    		else if(x==262143)
    		{
    			printf("87381
    ");
    		}
    		else if(x==524287)
    		{
    			printf("1
    ");
    		}
    		else if(x==1048575)
    		{
    			printf("349525
    ");
    		}
    		else if(x==2097151)
    		{
    			printf("299593
    ");
    		}
    		else if(x==4194303)
    		{
    			printf("1398101
    ");
    		}
    		else if(x==8388607)
    		{
    			printf("178481
    ");
    		}
    		else if(x==16777215)
    		{
    			printf("5592405
    ");
    		}
    		else if(x==33554431)
    		{
    			printf("1082401
    ");
    		}
            else
    		{
    		for(int t=1;t<=26;t++)
            {
            	sum*=2;
            	if(sum<=x&&sum*2>x)
            	{
            		sum*=2;
            		break;
    			}
    		}
    		cout<<sum-1<<endl;
    	   }
    	}
    	
    	return 0;
    }
    
  • 相关阅读:
    工作总结(二):Web Design
    工作总结(一):Linux C
    三十分钟学会AWK
    MySQL并发复制系列二:多线程复制 2016
    修改MySQL 5.7.9版本的root密码方法以及一些新变化整理
    sync_binlog innodb_flush_log_at_trx_commit 浅析
    MariaDB的"response time"插件
    Python学习九:列表生成式
    python中的深拷贝和浅拷贝理解
    Mycat 配置
  • 原文地址:https://www.cnblogs.com/Staceyacm/p/10781807.html
Copyright © 2011-2022 走看看