暑假集训7

A. One

考场忘了约瑟夫问题，以为是板子，记得有一个关于二进制的递推式，然后一直在打表，然后就没有然后了。。

这个不是我看过的那个约瑟夫。。。

考虑最开始有 $n$ 个人，现在场上还剩 $x$ 人，那么我要把第 $n - x + 1$ 个人踢走，踢走之后，第$n - x + 2$ 个人变成了剩余 $x - 1$ 个人时的第一个人，通过递归我们能得到幸存者在 $x - 1$ 个人的标号中的新标号 $id$，那么通过上面的关系，我们可以推出他在剩余 $x$ 个人时的标号为 $id + n - x +1$ ，回溯即可求得答案

但是不要写递归， $1e7$ 直接爆栈

开个数组倒着推回来就行了

code

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<random>
#include<cassert>
using namespace std;
const int maxn = 1e7 + 10;
inline int read(){
	int x = 0; char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
	return x;
}
int a[maxn];
int work(int n){
	a[1] = 1;
	for(int i = 2; i <= n; ++i){
		a[i] = ((n - i + 1) + a[i - 1]) % i;
		a[i] = a[i] ? a[i] : i;
	}
	return a[n];
}
int main(){
	int t = read();
	for(int i = 1; i <= t; ++i)printf("%d\n",work(read()));
	return 0;
}

B. 砖块

~~依题意模拟即可~~

考场只留 $20min$ 打这个题，果然锅了

输出for(int i = 1; i <= hi; ++i)printf("\%d ",dy + hi)

我是**

code

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<random>
#include<cassert>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int maxn = 105;
const int base = 1000;
int mp[2005][2005], hi, len;
char c[maxn];
void work(){
	int dx = 0, dy = 0, op = 1;
	mp[base][base] = 1;
	for(int i = 1; i <= len; ++i){
		if(c[i] == 'N'){
			if(op == 1){++dy; for(int k = 0; k < hi; ++k)++mp[base + dx][base + dy + k]; op = 2;}
			else if(op == 2){dy = dy + hi; ++mp[base + dx][base + dy];op = 1;}
			else if(op == 3){++dy; for(int k = 0; k < hi; ++k)++mp[base + dx + k][base + dy];}
		}
		if(c[i] == 'S'){
			if(op == 1){dy = dy - hi; for(int k = 0; k < hi; ++k)++mp[base + dx][base + dy + k]; op = 2;}
			else if(op == 2){--dy; ++mp[base + dx][base + dy];op = 1;}
			else if(op == 3){--dy; for(int k = 0; k < hi; ++k)++mp[base + dx + k][base + dy];}
		}
		if(c[i] == 'E'){
			if(op == 1){++dx; for(int k = 0; k < hi; ++k)++mp[base + dx + k][base + dy];op = 3;}
			else if(op == 2){++dx; for(int k = 0; k < hi; ++k)++mp[base + dx][base + dy + k];}
			else if(op == 3){dx = dx + hi; ++mp[base + dx][base + dy]; op = 1;}
		}
		if(c[i] == 'W'){
			if(op == 1){dx = dx - hi;for(int k = 0; k < hi; ++k)++mp[base + dx + k][base + dy]; op = 3;}
			else if(op == 2){--dx; for(int k = 0; k < hi; ++k)++mp[base + dx][base + dy + k];}
			else if(op == 3){--dx; ++mp[base + dx][base + dy]; op = 1;}
		}
	}
	int mx = 1;
	for(int i = 1; i <= 2000; ++i)
	  for(int j = 1; j <= 2000; ++j){
		  mx = max(mx, mp[i][j]);
		  mp[i][j] = 0;
	  }
	if(op == 1){
		printf("%d\n%d\n",dx,dy);
	}
	if(op == 2){
		for(int i = 1; i <= hi; ++i)printf("%d ",dx);printf("\n");
		for(int i = 1; i <= hi; ++i)printf("%d ",dy + i - 1);printf("\n");
	}
	if(op == 3){
		for(int i = 1; i <= hi; ++i)printf("%d ",dx + i - 1);printf("\n");
		for(int i = 1; i <= hi; ++i)printf("%d ",dy);printf("\n");
	}
	printf("%d\n",mx);
}
int main(){
	int t; scanf("%d",&t);
	for(int ask = 1; ask <= t; ++ask){
		scanf("%d%s",&hi,c + 1);
		len = strlen(c + 1);
		work();
	}
	return 0;
}

C. 数字

我以为是个结论题，发现没有结论后以为是个数位$DP$，结果他是数论题....

对下面内容存在关于知识点的疑惑，可以复习一下数论(无耻引流)

答案转化为求 $(n!)_{10} \mod 10^3$

不好求，把 $10^3$ 拆成 $2^3 \times 5^3$，求出答案后 $crt$ 合并

那么我们现在求

$(n!)_{10} \mod 5^3$

$(n!)_{10} \mod 2^3$

发现由于 $2$ 的倍数很多，所以 $n > 5$ 时候 $(n!)_{10} \mod 2^3$ 一直为 $0$

所以可以只求

$(n!)_{10} \mod 5^3$

用 $exlucas$ 中的方法我们可以求出 $(n!)_{5} \mod 5^3$

通过 $exlucas$ 中的方法还可以求出 $n!$ 有多少 $5$ 因子，记为 $c_5$

$(n!)_{10} = (n!)_{5} \times inv_2^{c_5}$

那么我们就能得到 $res = (n!)_{10} \mod 5^3$

$crt$ 合并一下就好了，

或者有偷懒写法，因为答案在 $1000$ 以内，我们可以直接枚举 $2^3$ 的倍数，选择 $\mod 5^3 = res$ 的，就是答案

code

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<random>
#include<cassert>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pii;
const ll maxn = 1005;
const int base = 10000;
const int block = 4;
inline ll read(){
	ll x = 0; char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
	return x;
}
char c[maxn];
struct big{
	int a[maxn];
	big(){clear();}
	void clear(){memset(a, 0, sizeof(a));}
	void in(){
		clear();
		scanf("%s",c + 1); 
		int len = strlen(c + 1);
		a[0] = (len + block - 1) / block;
		for(int i = 1; i <= len; ++i){
			int j = (len - i + block) / block;
			a[j] = a[j] * 10 + (c[i] ^ 48);
		}
		while(a[a[0]] == 0 && a[0]) --a[0];
	}
	friend big operator / (big a, const int x){
		int res = 0;
		for(int i = a.a[0]; i > 0; --i){
			res = res * base + a.a[i];
			a.a[i] = res / x; res %= x;
		}
		while(!a.a[a.a[0]] && a.a[0]) --a.a[0];
		return a;
	}
	int operator % (const int x)const{
		int res = 0;
		for(int i = a[0]; i > 0; --i){
			res = res * base + a[i];
			res %= x;
		}
		return res;
	}
}a;
int qpow(int x, int y, int mod){
	int ans = 1;
	for(; y; y >>= 1, x = x * x % mod)if(y & 1)ans = ans * x % mod;
	return ans;
}
int get_f(big now, int p, int pk){
	if(now.a[0] == 0)return 1;
	int res = 1, ls = now % pk, num;
	for(int i = 2; i <= ls; ++i)if(i % p)res = res * i % pk;
	num = res;
	for(int i = ls + 1; i < pk; ++i)if(i % p)res = res * i % pk;
	return get_f(now / p, p, pk) * qpow(res, (now / pk) % 100, pk) % pk * num % pk;
}
int getc5(){
	int c5 = 0; big b; b = a;
	while(b.a[0]){c5 = (c5 + (b / 5) % 100) % 100; b = b / 5;}
	return c5;
}
int work(){
	if(a.a[0] == 1 && a.a[0] <= 5){
		int n = a.a[1], cnt = 0; ll ans = 1;
		for(int i = 2; i <= n; ++i){
			int now = i; while(now % 5 == 0)now /= 5, ++cnt;
			ans = ans * now;
			while(cnt && ans % 2 == 0)ans = ans / 2, --cnt;
			ans %= 100000;
		}
		return ans % 1000;
	}
	int c5 = getc5();
	int res = get_f(a, 5, 125);
	int mod125 = res * qpow(63, c5 % 100, 125) % 125;
	for(int i = 8; i <= 1000; i += 8)if(i % 125 == mod125)return i;
}
int main(){
	int t; scanf("%d",&t);
	for(int ask = 1; ask <= t; ++ask){
		a.in(); int k; scanf("%d",&k);
		if(k == 1)printf("%d\n",work() % 10);
		if(k == 2)printf("%02d\n",work() % 100);
		if(k == 3)printf("%03d\n",work() % 1000);
	}
	return 0;
}

D. 甜圈

~~分块大法好~~

对每个块维护连续合法操作 $ l_i, r_i$ 新加操作不是 $r_i +1$直接非法，改成 $-1$即可

对单点的修改也非常省心

~~比线段树强多了~~

code

#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<random>
#include<cassert>
#include<cmath>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
const int maxn = 200005;
const int maxb = 505;
inline int read(){
	int x = 0; char c = getchar();
	while(c < '0' || c > '9')c = getchar();
	do{x = (x << 3) + (x << 1) + (c ^ 48); c = getchar();}while(c <= '9' && c >= '0');
	return x;
}
int block[maxn], l[maxb], r[maxb], now[maxn], n, k, len;
void push_down(int blo){
	if(l[blo]){
		int pl = (blo - 1) * len + 1, pr = min(blo * len, n);
		for(int i = pl; i <= pr; ++i){
			if(now[i] + 1 == l[blo])now[i] = r[blo];
			else now[i] = -1;
		}
		l[blo] = r[blo] = 0;
	}
}
void modify_one(int ll, int rr, int x){
	for(int i = ll; i <= rr; ++i){
		if(now[i] + 1 == x)++now[i];
		else now[i] = -1;
	}
}
void solve(int ll, int rr, int x){
	push_down(block[ll]); push_down(block[rr]);
	if(block[ll] == block[rr]){modify_one(ll, rr, x);return;}
	modify_one(ll, min(block[ll] * len, n), x);
	modify_one((block[rr] - 1) * len + 1, rr, x);
	for(int i = block[ll] + 1; i < block[rr]; ++i){
		if(l[i] != -1){
			if(l[i] == 0)l[i] = r[i] = x;
			else{
				if(r[i] + 1 == x)++r[i];
				else l[i] = r[i] = -1;
			}
		}
	}
}
int main(){
	n = read(); k = read();
	len = sqrt(n);
	for(int i = 1; i <= n; ++i)block[i] = (i + len - 1) / len;
	int t = read();
	for(int ask = 1; ask <= t; ++ask){
		int ll = read(), rr = read(), x = read();
		solve(ll, rr, x);
	}
	for(int i = 1; i <= block[n]; ++i)push_down(i);
	int ans = 0;
	for(int i = 1; i <= n; ++i)if(now[i] == k)++ans;
	printf("%d\n",ans);
	return 0;
}

posted @ 2022-08-21 18:51 Chen_jr 阅读(29) 评论(0) 编辑收藏举报

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Chen_jr

暑假集训7

A. One

B. 砖块

C. 数字

D. 甜圈

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