Codeforces Round #774 (Div. 2)

比赛链接

Codeforces Round #774 (Div. 2)

C. Factorials and Powers of Two

A number is called powerful if it is a power of two or a factorial. In other words, the number $m$ is powerful if there exists a non-negative integer $d$ such that $m=2^{d}$ or $m=d !$, where $d !=1 \cdot 2 \cdot \ldots \cdot d$ (in particular, 0 ! $=1$ ). For example 1,4 , and 6 are powerful numbers, because $1=1 !, 4=2^{2}$, and $6=3 !$ but 7,10 , or 18 are not.

You are given a positive integer $n$. Find the minimum number $k$ such that $n$ can be represented as the sum of $k$ distinct powerful numbers, or say that there is no such $k$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $t(1 \leq t \leq 100)$. Description of the test cases follows.
A test case consists of only one line, containing one integer $n\left(1 \leq n \leq 10^{12}\right)$

Output

For each test case print the answer on a separate line.
If $n$ can not be represented as the sum of distinct powerful numbers, print $-1$.
Otherwise, print a single positive integer - the minimum possible value of $k$.

Example

input

4
7
11
240
17179869184

output

2
3
4
1

Note

In the first test case, 7 can be represented as $7=1+6$, where 1 and 6 are powerful numbers. Because 7 is not a powerful number, we know that the minimum possible value of $k$ in this case is $k=2$.
In the second test case, a possible way to represent 11 as the sum of three powerful numbers is $11=1+4+6$. We can show that there is no way to represent 11 as the sum of two or less powerful numbers.

In the third test case, 240 can be represented as $240=24+32+64+120$. Observe that $240=120+120$ is not a valid representation, because the powerful numbers have to be distinct.

In the fourth test case, $17179869184=2^{34}$, so 17179869184 is a powerful number and the minimum $k$ in this case is $k=1$.

解题思路

思维，dfs

可以发现不可能出现无解的情况（由二进制表示，一个数总是可以表示为若干个不相同的二次幂相加的形式），可以将所有小于等于 $1e12$ 的乘方表示出来，共有 $16383$ 个数，然后每次暴力每次选择某一个乘方数或不选，最后再用若干个二次幂来表示剩余的数，注意当乘方数中包含 $1$ 或 $2$ 时，剩余的数的二次幂不能出现 $1$ 或 $2$，否则两个数会重复

• 时间复杂度：$O(16383\times T)$

代码

// Problem: C. Factorials and Powers of Two
// Contest: Codeforces - Codeforces Round #774 (Div. 2)
// URL: https://codeforces.com/contest/1646/problem/C
// Memory Limit: 256 MB
// Time Limit: 3000 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;
}


int t,tot;
LL n,a[60];
vector<int> chosen;
unordered_map<LL,int> f;
unordered_set<LL> s1,s2;
void dfs(int x)
{
	if(x==tot+1)
	{
		LL t=0;
		bool fl[2]={0};
		for(int i:chosen)
		{
			if(a[i]==1)fl[0]=1;
			if(a[i]==2)fl[1]=1;
			t+=a[i];
		}
		if(t>1e12||t==0)return ;
		if(fl[0])s1.insert(t);
		if(fl[1])s2.insert(t);
		f[t]=chosen.size();
		return ;
	}
	chosen.pb(x);
	dfs(x+1);
	chosen.pop_back();
	dfs(x+1);
}
int main()
{
    LL b=1;
    for(int i=1;;i++)
    {
    	b*=i;
    	if(b>1e12)break;
    	a[++tot]=b;
    }
    dfs(1);
    for(cin>>t;t;t--)
    {
    	cin>>n;
    	int res=__builtin_popcountll(n);
    	for(auto t:f)
    	{
    		LL x=t.fi;
    		int y=t.se;
    		if(x>n)continue;
    		LL z=n-x;
    		if(s1.count(x)&&(z&1))continue;
    		if(s2.count(x)&&(z>>1&1))continue;
    		res=min(res,y+__builtin_popcountll(z));
    	}
    	cout<<res<<'\n';
    }
    return 0;
}

__EOF__

本文作者：acwing_zyy
本文链接：https://www.cnblogs.com/zyyun/p/16048290.html
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