[构造]triples I

最新推荐文章于 2019-08-30 21:48:00 发布

转载最新推荐文章于 2019-08-30 21:48:00 发布 · 180 阅读

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原文链接：http://www.cnblogs.com/lllxq/p/11256345.html

本文针对一个具体的编程问题，即通过位运算求解特定输入条件下输出的最小数量的多个3的倍数，使得这些数的按位或结果等于给定的目标数。文章详细介绍了算法思路，包括如何使用位运算和模运算来确定所需输入数的数量和具体数值，并提供了一段完整的C++代码实现。

链接：https://ac.nowcoder.com/acm/contest/884/D?&headNav=acm&headNav=acm

来源：牛客网

题目描述

Doctor Elephant is testing his new program: output the bitwise or of the numbers inputed.

He has decided to input several multiples of 3 and the output of the program should be his favorite number

Because he is lazy, he decided to input as few numbers as possible. He wants you to construct such an input for every

It's guaranteed that for every

输入描述:

There're multiple test cases in a test file.

The first line contains a positive integer

In each of the following

输出描述:

For each test case output a line. First you need to output the number of numbers in your input, and then you need to output these numbers, separating by spaces.

示例1

输入

复制

2
3
7

输出

复制

1 3
2 3 6

说明

3=3, (3|6)=7

备注:

 $\leq T \leq 10^51≤T≤105, 1≤a≤10181 \leq a \leq 10^{18}1≤a≤1018.思路：$

AC代码：

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;

vector<ll> v[5];

void init(ll a){
  ll now=1;
  while(a){
    if(a&1) v[now%3].push_back(now);
    now<<=1;
    a>>=1;
  }
}

int main()
{
    ll t;scanf("%lld",&t);
    while(t--){
    ll a;scanf("%lld",&a);
    for(ll i=0;i<3;i++) v[i].clear();
    if(a%3==0) printf("1 %lld\n",a);
    else{
        init(a);

        printf("2 ");
        if(a%3==1){
            if(v[1].size()>=2){
                ll p=v[1][0],q=v[1][1];
                printf("%lld %lld\n",a-p,a-q);
            }
            else if(v[1].size()==1){
                ll p=v[1][0],q=v[2][0];
                printf("%lld %lld\n",a-p,p+q);
            }
            else{
                ll p=v[2][0],q=v[2][1],r=v[2][2];
                printf("%lld %lld\n",a-p-q,p+q+r);
            }
        }else{
            if(v[2].size()>=2){
                ll p=v[2][0],q=v[2][1];
                printf("%lld %lld\n",a-p,a-q);
            }
            else if(v[2].size()==1){
                ll p=v[2][0],q=v[1][0];
                printf("%lld %lld\n",a-p,p+q);
            }
            else{
                ll p=v[1][0],q=v[1][1],r=v[1][2];
                printf("%lld %lld\n",a-p-q,p+q+r);
            }
        }
    }
    }
    return 0;
}