2022.9.4———HZOI【CSP-S模拟2(联考)】

$Write\ In\ Front$

$Rank33/43$ 只拿了 $13 + 25 + 32 + 7 = 77pts$

$lyin$大佬三百多分，我一百不到，我太菜了

$$\Huge \mathbf{水博客警告！}$$

这篇博客里基本没写啥

T1 谜之阶乘，T2 子集，T3混凝土粉末，T4排水系统

$\mathfrak{T1}\ 谜之阶乘$

蜜汁阶乘太草了

注意到a-b最大差值不会超过$20$，因为$20!$已经达到了10^18级别

既然这个差值这么小，那么就可以枚举这个差值，然后套个二分找$a_i\ b_i$

作总结

$$枚举差值delta = a-b，然后枚举a_i，得到b_i = a_i - delta，然后check()算出res，二分判断l、r左右移动$$

然后就切了

怕爆long long可以开__int128，__int128手写个输入输出就行

T1

#include <iostream>
#include <cctype>
#include <vector>
#define GMY (520&1314)
#define FBI_OPENTHEDOOR(x) freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout);
#define re register int
#define char_phi signed
#define MARK cout << "###"
#define MARKER "@@@"
#define LMARK "!!!~~~"
#define ZY " qwq "
#define _ ' '
#define Endl cout << '\n'
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define N 25
#define INF 1000000000000000000
using namespace std;
inline __int128 read(){
	__int128 x(0); char c;
	while (!isdigit(c = getchar()));
	do{
		x = (x << 3) + (x << 1) + (c & 15);
	}while (isdigit(c = getchar()));
	return x;
}
void ot(__int128 x){
	if (x > 9) ot(x/10);
	putchar(x%10+'0');
}
__int128 T, n;
struct Final_Ans{__int128 ai, bi;};
vector<Final_Ans> a;
inline __int128 check(__int128 delta, __int128 mid){
	__int128 res(1);
	for (__int128 i = mid ; i >= mid-delta+1 ; -- i){
		res = res * i;
		if (res > INF)
			return INF+1;
	}
	return res;
}
inline void Clean(){
	a.clear();
}
void work(){
	Clean();
	n = read();
	if (n == 1)
		{puts("-1"); return ;}
	for (__int128 i = 20, l, r, mid, ans ; i >= 1 ; -- i){
		l = i+1, r = INF, ans = INF+1;
		while (l <= r){
			mid = ((l+r) >> 1);
			if (check(i, mid) < n)
				l = mid+1;
			else if (check(i, mid) > n)
				r = mid-1;
			else 
				{ans = mid; break;}
		}
		if (ans <= INF)
			a.push_back((Final_Ans){ans, ans-i});
	}
	ot((__int128)a.size()), putchar('\n');
	for (re i = 0 ; i < a.size() ; ++ i){
		ot(a[i].ai), putchar(' '), ot(a[i].bi), putchar('\n');
	}
}
// #define IXINGMY
char_phi main(){
    #ifdef IXINGMY
        FBI_OPENTHEDOOR(a);
    #endif
	T = read();
	while (T --)
	    work();
    return GMY;
}

$\mathfrak{T2}\ 子集$

这玩意好玩

首先对于$\frac{n}{k}$为偶数就比较显然，小学奥数。就比如这个，$1\ 2\ 3\ |\ 4\ 5\ 6$，显然选$1+6，2+5，3+4$，类似的更大也是这么选

对于正解就是这个的拓展

但是我不是特别理解，所以贺了题解

等$dbx$理解了我再去烦他

T2

#include <iostream>
#include <vector>
#define GMY (520&1314)
#define FBI_OPENTHEDOOR(x) freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout);
#define re register int
#define char_phi signed
#define MARK cout << "###"
#define MARKER "@@@"
#define LMARK "!!!~~~"
#define ZY " qwq "
#define _ ' '
#define Endl cout << '\n'
#define MIN(x, y) (((x) < (y)) ? (x) : (y))
#define MAX(x, y) (((x) > (y)) ? (x) : (y))
#define N 1000005
using namespace std;
inline void Fastio_setup(){ios::sync_with_stdio(false); cin.tie(NULL), cout.tie(NULL), cerr.tie(NULL);}
int T, n, K, len;
vector<int> a[N];
inline void doi(int id, int kuainum){
	for (re i = 1 ; i <= kuainum ; ++ i){
		if ((i & 1) == 1)
			for (re j = 1 ; j <= K ; ++ j)
				a[j].push_back(++ id);
		else 
			for (re j = K ; j >= 1 ; -- j)
				a[j].push_back(++ id);
	}
}
inline void doit(){
	int id(0);
	for (re i = 1 ; i <= K ; ++ i)
		a[i].push_back(++ id);
	for (re i = (K/2)+2 ; i <= K ; ++ i)
		a[i].push_back(++ id);
	for (re i = 1 ; i <= (K/2)+1 ; ++ i)
		a[i].push_back(++ id);
	for (re i = 1 ; i <= K ; ++ i)
		a[i].push_back(3*(3*K+1)/2 - a[i][0] - a[i][1]);
}
void work(){
	// Clean();
	cin >> n >> K; len = n/K;
	if (K == 1){
		cout << "Yes" << '\n';
		for (re i = 1 ; i <= n ; ++ i)
			cout << i << _;
		Endl;
		return ;
	}
	if (((n & 1) == 0 and (len & 1) == 1) or len == 1)
		{cout << "No" << '\n'; return ;}
	for (re i = 1 ; i <= K ; ++ i)
		vector<int>().swap(a[i]);
	cout << "Yes" << '\n';
	if ((len & 1) == 0)
		doi(0, len);
	else 
		doit(), doi(K*3, len-3);
	for (re i = 1 ; i <= K ; ++ i){
		for (re j = 0 ; j < len ; ++ j)
			cout << a[i][j] << _;
		Endl;
	}
}
// #define IXINGMY
char_phi main(){
    #ifdef IXINGMY
        FBI_OPENTHEDOOR(a);
    #endif
    Fastio_setup();
    cin >> T;
    while (T --)
	    work();
    return GMY;
}

$\mathfrak{T3}\ 混凝土粉末$

我水

$\mathfrak{T4}\ 排水系统$

嘉晚饭是真的