Codeforces Round #760 (Div. 3)

比赛链接

Codeforces Round #760 (Div. 3)

E. Singers' Tour

$n$ towns are arranged in a circle sequentially. The towns are numbered from 1 to $n$ in clockwise order. In the $i$-th town, there lives a singer with a repertoire of $a_{i}$ minutes for each $i \in[1, n]$.

Each singer visited all $n$ towns in clockwise order, starting with the town he lives in, and gave exactly one concert in each town. In addition, in each town, the $i$-th singer got inspired and came up with a song that lasts $a_{i}$ minutes. The song was added to his repertoire so that he could perform it in the rest of the cities.

Hence, for the $i$-th singer, the concert in the $i$-th town will last $a_{i}$ minutes, in the $(i+1)$-th town the concert will last $2 \cdot a_{i}$ minutes, ..., in the $((i+k) \bmod n+1)$-th town the duration of the concert will be $(k+2) \cdot a_{i}, \ldots$, in the town $((i+n-2) \bmod n+1)-n \cdot a_{i}$ minutes.

You are given an array of $b$ integer numbers, where $b_{i}$ is the total duration of concerts in the $i$-th town. Reconstruct any correct sequence of positive integers $a$ or say that it is impossible.

Input

The first line contains one integer $t\left(1 \leq t \leq 10^{3}\right)$ - the number of test cases. Then the test cases follow.
Each test case consists of two lines. The first line contains a single integer $n\left(1 \leq n \leq 4 \cdot 10^{4}\right)-$ the number of cities. The second line contains $n$ integers $b_{1}, b_{2}, \ldots, b_{n}\left(1 \leq b_{i} \leq 10^{9}\right)$ - the total duration of concerts in $i$-th city.
The sum of $n$ over all test cases does not exceed $2 \cdot 10^{5}$.

Output

For each test case, print the answer as follows:
If there is no suitable sequence $a$, print NO. Otherwise, on the first line print YES, on the next line print the sequence $a_{1}, a_{2}, \ldots, a_{n}$ of $n$ integers, where $a_{i}\left(1 \leq a_{i} \leq 10^{9}\right)$ is the initial duration of repertoire of the $i$-th singer. If there are multiple answers, print any of them.

Example

input

4
3
12 16 14
1
1
3
1 2 3
6
81 75 75 93 93 87

output

YES
3 1 3 
YES
1 
NO
YES
5 5 4 1 4 5 

Note

Let's consider the 1-st test case of the example:

1. the 1-st singer in the 1-st city will give a concert for 3 minutes, in the 2 -nd - for 6 minutes, in the 3 -rd - for 9 minutes;
2. the 2 -nd singer in the 1 -st city will give a concert for 3 minutes, in the 2 -nd - for 1 minute, in the 3 -rd - for 2 minutes;
3. the 3 -rd singer in the 1 -st city will give a concert for 6 minutes, in the 2 -nd - for 9 minutes, in the 3 -rd - for 3 minutes.

解题思路

构造，线性代数

即求解如下 $n$ 个 $n$ 元一次方程组：
$\begin{cases} & a_1+ n\times a_2+(n-1)\times a_3+\dots +2\times a_n=b_1 \\ & 2\times a_1+a_2+n\times a_3+\dots +3\times a_n=b_2 \\ & \vdots \\ & n\times a_1 + (n-1)\times a_2 +(n-2)\times a_3+\dots +a_n=b_n \end{cases}$

当高斯消元的复杂度为 $n^3$，会超时，不妨先从小点的范围开始构造
构造答案前，将上面的等式加起来，有 $\frac{(n+1)\times n}{2} \times \sum_{i=1}^n a_i=\sum_{i=1}^n b_i$，则必须有 $\sum_{i=1}^n b_i \%\frac{(n+1)\times n}{2} =0$
从 $n=3$ 开始构造：
有如下等式：
$\begin{cases} & a_1+2a_2+2a_3=b_1 \\ & 2a_1+a_2+3a_3=b_2 \\ \end{cases}$
做差，得 $-a_1+2a_2-a_3=b_1-b_2$，即 $a_2=\frac{b_1-b_2+a_1+a_2+a_3}{3}$，
得构造得到解：$a[i\%n+1]=\frac{b[i]-b[i\%n+1]+\sum_{i=1}^na[i]}{n}$，注意 $a[i]$ 为正整数，且范围为 $[1,10^9]$

• 时间复杂度：$O(n)$

代码

// Problem: E. Singers' Tour
// Contest: Codeforces - Codeforces Round #760 (Div. 3)
// URL: https://codeforces.com/contest/1618/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

const int N=4e4+5;
int n,t;
LL a[N],b[N];
int main()
{
    for(scanf("%d",&t);t;t--)
    {
    	scanf("%d",&n);
    	LL sum=0;
    	for(int i=1;i<=n;i++)scanf("%lld",&b[i]),sum+=b[i];
    	
    	int all=1ll*n*(n+1)/2;
    	if(sum%all!=0)
    	{
    		puts("NO");
    		continue;
    	}
    	sum/=all;
    	bool f=true;
    	for(int i=1;i<=n;i++)
    	{
    		LL t=b[i]-b[i%n+1]+sum;
    		if(t%n!=0)
    		{
    			f=false;
    			break;
    		}
    		a[i%n+1]=t/n;
    		if(!(a[i%n+1]>=1&&a[i%n+1]<=1e9))
    		{
    			f=false;
    			break;
    		}
    	}
    	
    	if(!f)puts("NO");
    	else
    	{
    		puts("YES");
    		for(int i=1;i<=n;i++)printf("%lld ",a[i]);
    		puts("");
    	}
    }
    return 0;
}

F. Reverse

You are given two positive integers $x$ and $y$. You can perform the following operation with $x$ : write it in its binary form without leading zeros, add 0 or 1 to the right of it, reverse the binary form and turn it into a decimal number which is assigned as the new value of $x$.
For example:

• $34$ can be turned into $81$ via one operation: the binary form of $34$ is $100010$ , if you add $1$ , reverse it and remove leading zeros, you will get $1010001$ , which is the binary form of $81$ .
• $34$ can be turned into $17$ via one operation: the binary form of $34$ is $100010$, if you add $0$ , reverse it and remove leading zeros, you will get $10001$ , which is the binary form of $17$ .
• $81$ can be turned into $69$ via one operation: the binary form of $81$ is $1010001$ , if you add $0$ , reverse it and remove leading zeros, you will get $1000101$ , which is the binary form of $69$ .
• $34$ can be turned into $69$ via two operations: first you turn $34$ into $81$ and then $81$ into $69$ .

Your task is to find out whether $x$ can be turned into $y$ after a certain number of operations (possibly zero).

Input

The only line of the input contains two integers $x$ and $y\left(1 \leq x, y \leq 10^{18}\right)$.

Output

Print YES if you can make $x$ equal to $y$ and NO if you can't.

解题思路

dfs，记忆化搜索

对于一个数二进制，如果其后面没有 $0$，则如果在其后面加 $0$ 的话翻转后前导 $0$ 会去掉，相当于翻转原二进制串，即后面只能加 $1$，这就相当于只能在原二进制串前后加 $1$，且可翻转；如果一个数的二进制后面有 $0$，则有两种选择：将后面的 $0$ 去掉，在后面加 $1$，这时两遍 $dfs$ 即可

• 时间复杂度：$O(32\times 32)$

代码

// Problem: F. Reverse
// Contest: Codeforces - Codeforces Round #760 (Div. 3)
// URL: https://codeforces.com/contest/1618/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

// %%%Skyqwq
#include <bits/stdc++.h>
 
//#define int long long
#define help {cin.tie(NULL); cout.tie(NULL);}
#define pb push_back
#define fi first
#define se second
#define mkp make_pair
using namespace std;
 
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
 
template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }
 
template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char s = getchar();
    while (s < '0' || s > '9') { if (s == '-') f = -1; s = getchar(); }
    while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
    x *= f;
}

LL x,y;
unordered_set<string> s;
string get(LL x)
{
	string res;
	do
	{
		res+='0'+(x&1);
		x>>=1;
	}while(x);
	return res;
}
void dfs(string x)
{
	if(s.count(x)||x.size()>=64)return ;
	s.insert(x);
	reverse(x.begin(),x.end());
	dfs(x);
	x+='1';
	dfs(x);
	x.pop_back();
	x='1'+x;
	dfs(x);
}
int main()
{
	cin>>x>>y;
	string st=get(x),en=get(y);
	reverse(st.begin(),st.end());
	reverse(en.begin(),en.end());
	if(st.back()=='0')
	{
		s.insert(st);
		dfs(st+'1');
		while(st.size()&&st.back()=='0')st.pop_back();
		dfs(st);
	}
	else
		dfs(st);
    puts(s.count(en)?"YES":"NO");
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/16250728.html
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