test20181017 B君的第一题

题意

分析

考场做法

对p的幂打表发现，我们一定可以把x和y的二进制位从低到高依次调整成0。

具体而言，从0次幂开始每两个分为一组a,b，那么0,a,b,a+b组合中的一种可以将x,y的对应二进制位都调整成0。

然后模拟一下就行了。

时间复杂度\(O(\log |x| + \log |y|)\)

#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<complex>
#define rg register
#define il inline
#define co const
#pragma GCC optimize ("O0")
using namespace std;
template<class T> il T read(T&x)
{
    T data=0;
	int w=1;
    char ch=getchar();
    while(!isdigit(ch))
    {
		if(ch=='-')
			w=-1;
		ch=getchar();
	}
    while(isdigit(ch))
        data=10*data+ch-'0',ch=getchar();
    return x=data*w;
}
typedef long long ll;
const int INF=0x7fffffff;
const complex<ll>p(-1,-1);

const int MAXN=1000;
int S[MAXN],cnt;

int main()
{
  freopen("guangzhou.in","r",stdin);
  freopen("guangzhou.out","w",stdout);
	ll x,y;
	read(x);read(y);
	complex<ll>a(1,0),b(-1,-1),t;
	for(int i = 0;x || y;++i,(a *= p) *= p,(b *= p) *= p)
	{
//		cerr<<"i="<<i<<endl;
//		cerr<<"a="<<a<<" b="<<b<<endl;
		t = 0;
		if( (x & (1LL << i)) == (t.real() & (1LL << i)) && (y & (1LL << i)) == (t.imag() & (1LL << i)) )
		{
//			cerr<<" case 1"<<endl;
			continue;
		}
		t = a;
		if( (x & (1LL << i)) == (t.real() & (1LL << i)) && (y & (1LL << i)) == (t.imag() & (1LL << i)) )
		{
//			cerr<<" case 2"<<endl;
			x -= t.real() ,y -= t.imag();
			S[++cnt] = 2 * i;
			continue;
		}
		t = b;
		if( (x & (1LL << i)) == (t.real() & (1LL << i)) && (y & (1LL << i)) == (t.imag() & (1LL << i)) )
		{
//			cerr<<" case 3"<<endl;
			x -= t.real() ,y -= t.imag();
			S[++cnt] = 2 * i + 1;
			continue;
		}
		t = a + b;
		if( (x & (1LL << i)) == (t.real() & (1LL << i)) && (y & (1LL << i)) == (t.imag() & (1LL << i)) )
		{
//			cerr<<" case 4"<<endl;
			x -= t.real() ,y -= t.imag();
			S[++cnt] = 2 * i;
			S[++cnt] = 2 * i + 1;
			continue;
		}
	}
	printf("%d\n",cnt);
	for(int i=1;i<=cnt;++i)
	{
		printf("%d\n",S[i]);
	}
//  fclose(stdin);
//  fclose(stdout);
    return 0;
}

标解

跟冬令营2017亿兆京垓 (Radixphi)这道题有关。

像确定二进制一样，每次右移，然后判断最后一位的奇偶，这题可以每次/p，然后判断实部和虚部的和的奇偶。

高斯整数一定能表示成\(-1 \pm i\)进制的形式，这是B君翻维基百科上翻到的。然后就被出成题了。

时间复杂度\(O(\log |x| + \log |y|)\)。

#include <bits/stdc++.h>
using namespace std;
complex<long long> n, p, u;
long long x, y;
int a[200], c, i;
int main() {
	freopen("guangzhou.in", "r", stdin);
	freopen("guangzhou.out", "w", stdout);
	cin >> x >> y;
	n = complex<long long>(x, y);
	p = complex<long long>(-1, -1);
	while (n != complex<long long>(0, 0)) {
		if ((n.real() + n.imag()) % 2 != 0) {
			a[c++] = i;
			n -= complex<long long>(1, 0);
		}
		n /= p;
		i++;
	}
	printf("%d\n", c);
	for (int i = 0; i < c; i++) {
		printf("%d\n", a[i]);
	}
	return 0;
}