JZOJ 2020.08.03【NOIP提高组】模拟 &&【NOIP2015模拟11.5】

总结

又是一日爆炸
(T1) 不出所料报 (0) 了？！

题目

(T1)

JZOJ 4315. Prime

暴力就好了？！
考场根本没想暴力
赛后发现暴力跑得贼快
只需二分一下组数的上界
然后 (dfs) 判断能否能成功分完组
跑时顺便统计答案就行了

(Code)

#include<cstdio>
#include<iostream>
using namespace std;

const int N = 20;
int n , a[N] , vis[N][N] , d[N][N] , cnt , ans , Mx , bz;

inline int gcd(int x , int y){return y == 0 ? x : gcd(y , x % y);}

inline void dfs(int x , int mid , int Max)
{
	if (mid > ans || mid == ans && Max >= Mx) return;
	if (cnt > mid) return;
	if (x > n)
	{
		bz = 1;
		if (ans > mid) ans = mid , Mx = Max;
		else if (ans == mid && Max < Mx) Mx = Max;
		return;
	}
	for(register int i = 1; i <= cnt; i++)
	{
		int fl = 0;	
		for(register int j = 1; j <= d[i][0]; j++)
		if (!vis[x][d[i][j]]) 
		{
			fl = 1;
			break;
		}
		if (fl) continue;
		d[i][++d[i][0]] = x;
		dfs(x + 1 , mid , max(Max , d[i][0]));
		--d[i][0];
	}
	d[++cnt][++d[cnt][0]] = x;
	dfs(x + 1 , mid , max(Max , 1));
	--d[cnt][0] , --cnt;
}

int main()
{
	freopen("prime.in" , "r" , stdin);
	freopen("prime.out" , "w" , stdout);
	scanf("%d" , &n);
	for(register int i = 1; i <= n; i++) scanf("%d" , &a[i]);
	for(register int i = 1; i <= n; i++)
		for(register int j = 1; j <= n; j++)
		if (i != j) vis[i][j] = gcd(a[i] , a[j]) == 1 ? 1 : 0;
	ans = 0x3f3f3f3f , Mx = 0x3f3f3f3f;
	int l = 1 , r = n , mid;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		cnt = bz = 0;
		dfs(1 , mid , 0);
		if (bz) r = mid - 1;
		else l = mid + 1;
	}
	printf("%d %d" , ans , Mx);
}

(T2)

JZOJ 4316. Isfind

一眼没看出？！序列自动机？！？
去一边，暴力又能过？！！
天！！！
而我想到了非暴力的解法，幸好过了，不然亏大了
只需记录每种字母在原串出现的先后位置
然后匹配时二分找位置判断就行了

(Code)

#include<cstdio>
#include<cstring>
using namespace std;

const int N = 1e5 + 5;
int n , m , a[30][N] , p[30];
char s[N];

inline int binary(int t , int x)
{
	int l = 0 , r = p[t] , mid , res = -1;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (a[t][mid] >= x) res = a[t][mid] , r = mid - 1;
		else l = mid + 1;
	}
	return res;
}

int main()
{
	freopen("isfind.in" , "r" , stdin);
	freopen("isfind.out" , "w" , stdout);
	scanf("%d%d%s" , &n , &m , s);
	int len = strlen(s) , pos , pos1 , fl;
	for(register int i = 0; i < 28; i++) p[i] = -1;
	for(register int i = 0; i < len; i++) a[s[i] - 'a'][++p[s[i] - 'a']] = i;
	while (m--)
	{
		scanf("%s" , s);
		len = strlen(s);
		pos = -1;
		fl = 0;
		for(register int i = 0; i < len; i++)
		{
			pos1 = binary(s[i] - 'a' , pos + 1);
			if (pos1 == -1)
			{
				printf("N
");
				fl = 1;
				break;
			}
			else pos = pos1;
		}
		if (!fl) printf("Y
");
	}
}

实际上，它是序列自动机的模板题
所以上个序列自动机的代码

(Code)

#include<cstdio>
#include<cstring>
using namespace std;

const int N = 1e5 + 5 , INF = 0x3f3f3f3f;
int n , m , nxt[N][30];
char s[N];

int main()
{
	freopen("isfind.in" , "r" , stdin);
	freopen("isfind.out" , "w" , stdout);
	scanf("%d%d%s" , &n , &m , s);
	int len = strlen(s);
	for(register int i = 0; i <= 26; i++) nxt[len][i] = INF;
	for(register int i = len - 1; i >= 0; i--)
	{
		for(register int j = 0; j <= 26; j++) nxt[i][j] = nxt[i + 1][j];
		nxt[i][s[i] - 'a'] = i;
	}
	for(; m; --m)
	{
		scanf("%s" , s);
		len = strlen(s);
		int pos = -1 , fl = 0;
		for(register int i = 0; i < len; i++)
		{
			pos = nxt[pos + 1][s[i] - 'a'];
			if (pos == INF)
			{
				printf("N
") , fl = 1;
				break;
			}
		}
		if (!fl) printf("Y
");
	}
}

(T3)

JZOJ 4317. Divide

很显然 (a_i) 有用的部分是 (gcd(a_i,p))
然后我们就发现 (a_i imes a_j imes a_k) 相当于 (p) 的因数相乘
我们只要处理出 (p) 的所有因数，然后 (O(tot^3)) 枚举三个因数相乘
用桶记下每种因数在 (a) 出现的次数
然后分类讨论算贡献即可

(Code)

#include<cstdio>
using namespace std;
typedef long long LL;

const int N = 3e4 + 5 , M = 1e6 + 5;
LL a[N] , pr[M] , buc[M] , p , ans;
int n , tot;

inline LL gcd(LL x , LL y){return y == 0 ? x : gcd(y , x % y);}

int main()
{
	freopen("divide.in" , "r" , stdin);
	freopen("divide.out" , "w" , stdout);
	scanf("%d%lld" , &n , &p);
	for(register int i = 1; i <= n; i++)
	{
		scanf("%lld" , &a[i]);
		a[i] = gcd(a[i] , p);
		++buc[(int)a[i]];
	}	
	for(register int i = 1; i <= p; i++)
	if (p % i == 0) pr[++tot] = i;
	for(register int i = 1; i <= tot; i++)
	for(register int j = i; j <= tot; j++)
	for(register int k = j; k <= tot; k++)
	if (pr[i] * pr[j] % p * pr[k] % p == 0)
	{
		if (i == j && i == k) ans += buc[pr[i]] * (buc[pr[i]] - 1) * (buc[pr[i]] - 2) / 6;
		else{
			if (i == j) ans += buc[pr[i]] * (buc[pr[j]] - 1) * buc[pr[k]] / 2;
			else if (i == k) ans += buc[pr[i]] * (buc[pr[k]] - 1) * buc[pr[j]] / 2;
			else if (j == k) ans += buc[pr[j]] * (buc[pr[k]] - 1) * buc[pr[i]] / 2;
			else ans += buc[pr[i]] * buc[pr[j]] * buc[pr[k]];
		}
	}
	printf("%lld" , ans);
}