钙 模拟赛总结与代码

目录

• 10.28上午QBXT
• 10.27ZR
• 10.24
• 10.23
• 10.18
• 10.16
• 10.15
• 10.13
• 10.5
• 10.6
• 10.11普及模拟赛

10.28上午QBXT

比较简单的一套模拟赛吧。
T1：随便做

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second

inline int gi() {
  int x = 0,f = 1;char c = gc;
  while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
  return x * f;
}

const int maxN = 1e5 + 7;
char s[maxN];

int main() {
	freopen("increase.in","r",stdin);
	freopen("increase.out","w",stdout);
	scanf("%s",s + 1);
	int n = strlen(s + 1);
	int now = s[1] - '0';
	int pos = 1,tmp = 1;
	for(int i = 2;i <= n;++ i) {
		int x = s[i] - '0';
		if(x > now) {
			now = x;
			pos = i;
		}else {
			if(x < now) break;
		}
		if(x == now) tmp = i;
	}	
	if(tmp == n) {
		printf("%s",s + 1);
		return 0;
	}
	bool flag = false;
	for(int i = 1;i < pos;++ i) {
		printf("%c",s[i]); 
	}
	if(s[pos] - '0' - 1 != 0)
	printf("%c",s[pos] - 1);
	for(int i = pos + 1;i <= n;++ i) {
		printf("9");
	}
	return 0;
}

T2：随便做

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define int long long
const int mod = 1e12 + 7;

inline int gi() {
  int x = 0,f = 1;char c = gc;
  while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
  while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
  return x * f;
}

const int maxN = 4e4 + 7;

int a[maxN] , ans,p[100007],b[maxN],s[maxN];

int query(int x) {
	int ans = 0;
	for(;x;x -= x & -x) ans = (ans + p[x]) % mod;
	return ans;
}

const int MAX = 1e5;

void add(int x , int val) {
	for(;x <= MAX;x += x & -x) {
		p[x] = (p[x] + val) % mod;
	}
	return ;
}

int mul(int a , int b) {
	int ans = 0;
	for(int now = a;b;b >>= 1,now = (now + now) % mod) {
		if(b & 1) ans = (ans + now) % mod;
	}
	return ans;
}

signed main() {
	freopen("multiplication.in","r",stdin);
	freopen("multiplication.out","w",stdout);
	int n = gi();
	for(int i = 1;i <= n;++ i) s[i] = b[i] = gi();
	sort(s + 1,s + n + 1);
	for(int i = 1;i <= n;++ i) a[i] = lower_bound(s + 1,s + n + 1,b[i]) - s;
	for(int i = 1;i <= n;++ i) {
		int tmp = query(n) - query(a[i]);
		tmp %= mod;
		tmp += mod;
		tmp %= mod;
		ans = ans + mul(mul(tmp , n - i + 1) , b[i]);
		ans %= mod;
		add(a[i] , b[i] * i % mod);
	}
	ans %= mod;
	ans += mod;
	ans %= mod;
	cout << ans;
	return 0;
}

T3：简单题
nmd,我不会扫描线。
会了扫描线这是一个简单题吧
不过这里没有必要用线段树维护扫描线。
考虑出现

10.27ZR

T1

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
const int mod = 987654321;


inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int maxN = 1000 + 7;

int f[maxN], sum[maxN], a[maxN];
bool vis[1000007];
int prime[100007];
int num;

void Pre() {
	int n = 1000000;
	vis[1] = 1;
	for(int i = 2;i <= n;++ i) {
		if(!vis[i]) prime[++ num] = i;
		for(int j = 1;j <= num && i * prime[j] <= n;++ j) {
			vis[i * prime[j]] = true;
			if(i % prime[j] == 0) break;
		}
	}
	return;
}

int main() {
	Pre();
	int n = gi();
	for(int i = 1;i <= n;++ i) {
		a[i] = gi();
		sum[i] = sum[i - 1] + a[i];	
	}
	f[0] = 1;
	for(int i = 1;i <= n;++ i) {
		for(int j = 0;j < i;++ j) {
			if(!vis[sum[i] - sum[j]]) f[i] = (f[i] + f[j]) % mod;
		}
	}
	cout << f[n];
	return 0;
}

T2:

#include <bits/stdc++.h>
using namespace std;
const int maxN = 100000 + 7;
const int inf = 0x7fffffff;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}


int n, m;
int fa[maxN];

struct Edge{
	int u , v, w;
}a[maxN];

bool cmp(Edge a , Edge b) {
	return a.w > b.w;
}

int get(int x) {return fa[x] == x ? x : fa[x] = get(fa[x]);}
int merge(int x , int y) {fa[get(x)] = get(y);}

struct Node {
	int v , nex, w;
}Map[maxN];
int num , head[maxN];

void add_Node(int u , int v, int w) {
	Map[++ num] = (Node) {v , head[u], w};
	head[u] = num;
}

void Work_one() {
	sort(a + 1, a + m + 1, cmp);
	int cnt = 0;
	for(int i = 1;i <= m;++ i) {
		int fax = get(a[i].u) , fay = get(a[i].v);
		if(fax != fay) {
			merge(a[i].u , a[i].v);
			add_Node(a[i].u , a[i].v, a[i].w);add_Node(a[i].v , a[i].u, a[i].w);
			cnt ++;
			if(cnt == n - 1) break;
		}
	}
	return;
}

int dep[maxN];
int f[maxN][23], v[maxN][23];
bool vis[maxN];

void dfs(int now , int fa, int w) {
	dep[now] = dep[fa] + 1;
	vis[now] = true;
	f[now][0] = fa;
	v[now][0] = w;
	for(int i = 1;i <= 20;++ i) {
		f[now][i] = f[f[now][i - 1]][i - 1];
		v[now][i] = min(v[now][i - 1] , v[f[now][i - 1]][i - 1]);
	}
	for(int i = head[now];i;i = Map[i].nex) {
		int v = Map[i].v;
		if(v == fa) continue;
		dfs(v , now, Map[i].w);
	}
	return;
}

void Work_two() {
	for(int i = 1;i <= n;++ i) {
		if(!vis[i]) {
			dfs(i , 0, -1);
		}
	}
	
}

int lca(int x , int y) {
	int ans = inf;
	if(dep[x] > dep[y]) swap(x , y);
	for(int i = 20;i >= 0;-- i) {
		if(dep[y] - (1 << i) >= dep[x]) {ans = min(ans , v[y][i]);y = f[y][i]; }
	}
	if(x == y) return ans;
	for(int i = 20;i >= 0;-- i) {
		if(f[y][i] != f[x][i]) {
			ans = min(ans , v[y][i]);
			ans = min(ans , v[x][i]);
			y = f[y][i];
			x = f[x][i];
		}
	}
	ans = min(ans, min(v[x][0] , v[y][0]));
	return ans;
}

int main() {
//	freopen("testdata.in","r",stdin);
	n = gi(), m = gi();	int q = gi();
	rep(i , 1, n) fa[i] = i;
	for(int i = 1;i <= m;++ i) {a[i].u = gi();a[i].v = gi();a[i].w = gi();}
	Work_one();
	Work_two();

	for(int i = 1;i <= q;++ i) {
		int x = gi(), y = gi(), w = gi();
		int ans = lca(x , y);
		if(ans == -1) puts("No");
		else {
			if(ans >= w) {
				puts("Yes");
			}else puts("No");
		}
//		printf("%d\n",ans);
	}
	return 0;
}

T3

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define pz putchar('0');
#include <map>
using namespace std;
const int MO = 15;
struct Big {
	int len, data[10001];
	void clear() {
		memset(this, 0, sizeof(*this));
	}
	int & operator [] (int k) {
		return data[k];
	}
	Big & operator = (int k) {
		clear();
		len = 0;
		while(k) {
			++len;
			data[len] = k & MO;
			k >>= 4;
		}
		if (len == 0) ++len;
		return *this;
	}
	Big operator * (Big & A) {
		Big temp;
		temp.clear();
		temp.len = len + A.len - 1;
		for (int i = 1; i <= len; i++)
			for (int j = 1; j <= A.len; j++) {
				temp[i + j - 1] += A[j] * data[i];
				temp[i + j] += (temp[i + j - 1] >> 4);
				temp[i + j - 1] &= MO;
			}
		while(temp[temp.len + 1]) ++temp.len;
		return temp;
	}
	void print() {
		for (int i = len; i >= 1; i--) printf("%X", data[i]);
		putchar('\n');
	}
} temp, ans;
map<int, bool> M;
bool f[1000001];
int pnum, p[100001];
void GETP(int M) {
	memset(f, 1, sizeof(f));
	f[0] = f[1] = false;
	p[pnum = 1] = 2;
	for (int now = 2; now < M;) {
		for (int j = now + now; j <= M; j += now) f[j] = false;
		++now;
		while(now < M && !f[now]) ++now;
		if (f[now]) p[++pnum] = now;
	}
}
int gcd(int a, int b) {
	return b == 0 ? a : gcd(b, a % b);
}
void work(int num) {
	for (int i = 1; i <= pnum; i++) {
		if (num % p[i] == 0)
			if (M[p[i]] == 0) {
				M[p[i]] = true;
				temp = p[i];
				ans = ans * temp;
			}
		while(num % p[i] == 0) num /= p[i];
	}
	if (num != 1)
		if (M[num] == 0) {
			M[num] = true;
			temp = num;
			ans = ans * temp;
		}
}
int main() {
	ans = 1;
	int t;
	scanf("%d", &t);
	GETP(100000);
	while(t--) {
		int a, b;
		scanf("%d%d", &a, &b);
		int d = gcd(a, b);
		a /= d;
		b /= d;
		work(b);
	}
	ans.print();
}

10.24

T3不会做，T1、T2傻逼题
T1:

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
const int maxN = 1e7 + 7;
const int mod = 1e9 + 7;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

int f[2][2];
char s[maxN];

int main() {
//	freopen("zero.in","r",stdin);
//	freopen("zero.out","w",stdout);
	int n = gi();
	scanf("%s",s + 1);
	f[0][0] = 1;
	f[0][1] = 1;
	for(int i = 1;i <= n;++ i) {
		int now = i & 1;
		int last = now ^ 1;
		if(s[i] == '&') {
			f[now][0] = (1ll * f[last][0] * 2 + f[last][1]) % mod;
			f[now][1] = f[last][1];
		}
		if(s[i] == '|') {
			f[now][0] = f[last][0];
			f[now][1] = (1ll * f[last][1] * 2 + f[last][0]) % mod;
		}
		if(s[i] == '^') {
			f[now][0] = (f[last][1] + f[last][0]) % mod;
			f[now][1] = (f[last][0] + f[last][1]) % mod;
		}
	}
	printf("%d", f[n & 1][1]);
	return 0;
}

T2：

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define int long long
const int mod = 1e9 + 7;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

int gcd(int a , int b) {
	return !b ? a : gcd(b , a % b);
}

int L, R, K;
void Sub1() {
	int ans = 0;
	for(int a = L;a <= R - 2 * K;++ a) {
		if(gcd(a , 2 * K) == 1) {
			ans ++;
		}
	}
	printf("%d\n",ans);
}

vector<int> q;
int n;

void work(int x) {
	int p = sqrt(x);
	for(int i = 2;i <= p;++ i) {
		if(x % i == 0) {
			q.push_back(i);
			++ n;
			while(x % i == 0) x /= i;
		}
	}
	if(x != 1) {
		q.push_back(x);
		++ n; 
	} 
	return;
}

int sum;
bool vis[312];
int TQL;

void dfs(int now , int tot, int pi) {
	if(now == n + 1) {
		if(tot & 1) sum -= TQL / pi;
		else sum += TQL / pi;
		return;
	}
	dfs(now + 1,tot + 1, pi * q[now - 1]);
	dfs(now + 1,tot, pi);
}

int solve() {
	sum = 0;
	dfs(1 , 0, 1LL);
	return sum;
}

signed main() {
//	freopen("prime.in","r",stdin);
//	freopen("prime.out","w",stdout);
	L = gi(), R = gi(),K = gi();
	if(R - L <= 1000000) return Sub1(),0;
	work(2 * K);
	TQL = R - 2 * K;
	int ans = solve();
	TQL = L - 1;
	ans -= solve();
	cout << ans % mod;
	return 0;
}

10.23

懒得讲，都挺简单的。
T1：

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define int long long
const int mod = 998244353;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

int Gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = (1ll * x * 10 % (mod - 1) + c - '0') % (mod - 1);c = gc;}
	return x * f;
}

const int maxN = 1e6 + 7;

char s[maxN];
int fast_pow(int a, int b) {
	int ans = 1;
	for(int now = a;b;b >>= 1,now = now * now % mod) {
		if(b & 1) ans = ans * now % mod;
	}
	return ans;
}

int inv(int x) {
	return fast_pow(x , mod - 2);
}

int n;

signed main() {
//	freopen("a.in","r",stdin);
//	freopen("a.out","w",stdout);
//	cin >> n;
	int n = Gi();
	scanf("%s",s + 1);
	int l = strlen(s + 1);
	if(l > n)  return cout << fast_pow(26ll,n)<<"\n",0;
	else if(l == n) return cout << fast_pow(26ll,n) - 1,0;
	int ans = fast_pow(26ll,n);
	ans = ans - (25ll * l % mod) * fast_pow(26ll , n) % mod * inv(fast_pow(26ll,l + 1)) % mod;
	ans = ans - fast_pow(26,n - l) ;
	ans %= mod;
	ans += mod;
	ans %= mod;
	cout << ans;
	return 0;
}
/*
10000000000 aaa
*/

T2：

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define lson now << 1
#define rson now << 1 | 1
#define int long long

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int maxN = 1e5 + 7;
struct Node {
	int x , y;
}a[maxN];

int f[maxN];
int b[maxN << 1];
PII p[maxN];

struct Tree {
	int l , r;
	int maxx;
}tree1[maxN << 3] , tree2[maxN << 3];

void updata(int now , int id) {
	if(id == 1) tree1[now].maxx = max(tree1[lson].maxx , tree1[rson].maxx);
	else tree2[now].maxx = max(tree2[lson].maxx , tree2[rson].maxx);
}

void build(int now , int l, int r) {
	tree1[now].l = l;tree1[now].r = r;
	tree2[now].l = l;tree2[now].r = r;
	if(l == r) return;
	int mid = (l + r) >> 1;
	build(lson , l, mid);
	build(rson , mid + 1, r);
}

void change(int now , int id, int pos, int val) {
	if(tree1[now].l == tree1[now].r) {
		if(id == 1) tree1[now].maxx = max(tree1[now].maxx , val);
		else tree2[now].maxx = max(tree2[now].maxx , val);
		return ;
	}
	int mid = (tree1[now].l + tree1[now].r) >> 1;
	if(pos <= mid) change(lson , id,pos,val);
	else change(rson , id,pos,val);
	updata(now , id);
}

int query(int now , int l , int r , int id) {
	if(tree1[now].l >= l && tree1[now].r <= r) {
		if(id == 1)  return tree1[now].maxx;
		else return tree2[now].maxx;
	}
	int mid = (tree1[now].l + tree1[now].r) >> 1 , ans = 0;
	if(l <= mid) ans = max(ans , query(lson , l, r, id));
	if(r > mid) ans = max(ans , query(rson , l, r, id));
	return ans;
}

int num;

signed main() {
//	freopen("b.in","r",stdin);
//	freopen("b.out","w",stdout);
	int n = gi();
	rep(i , 1, n) {
		a[i].x = gi();
		a[i].y = gi();
		b[++ num] = a[i].x;
		b[++ num] = a[i].y;
	}
	sort(b + 1,b + num + 1);
	for(int i = 1;i <= n;++ i) {
		p[i].fi = lower_bound(b + 1,b + num + 1,a[i].x) - b;
		p[i].se = lower_bound(b + 1,b + num + 1,a[i].y) - b;
	}
	build(1 , 1, num);
	for(int i = 1;i <= n;++ i) {
		if(a[i].x == a[i].y) continue;
		f[i] = 1;
		if(a[i].x > a[i].y) {
			f[i] = max(f[i] , query(1 , 1, p[i].fi - 1, 1) + 1);
			change(1 , 1, p[i].se, f[i]);
//			printf("%d",)
		}else {
			f[i] = max(f[i] , query(1 , p[i].fi + 1, num, 2) + 1);
			change(1 , 2, p[i].se, f[i]);
		}
	}
	int ans = 0;
	for(int i = 1;i <= n;++ i) ans = max(ans , f[i]);
	printf("%d",ans);
	return 0;
}

T3：

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
const int mod = 998244353;
const int maxN = 1e6 + 7;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

struct Edge {
	int u , v;
}a[maxN];

struct Node {
	int v , nex;
}Map[maxN << 1];
int head[maxN] , num;
PII b[maxN];

void add_Node(int u , int v) {
	Map[++ num] = (Node) {v , head[u]};
	head[u] = num;
}

int dep[maxN] , f[maxN][23], fa[maxN];

void dfs(int now,int fa) {
	dep[now] = dep[fa] + 1;
	f[now][0] = fa;
	for(int i = 1;i <= 21;++ i) f[now][i] = f[f[now][i - 1]][i - 1];
	for(int i = head[now];i;i = Map[i].nex) {
		int v = Map[i].v;
		if(v == fa) continue;
		dfs(v , now);
	}
}

int lca(int x , int y) {
	int ans = 0;
	if(dep[x] > dep[y]) swap(x , y);
	for(int i = 21;i >= 0;-- i) {
		if(dep[f[y][i]] >= dep[x]) {
			y = f[y][i];
			ans += 1 << i;
		} 
	}
	if(x == y) return ans;
	for(int i = 21;i >= 0;-- i) {
		if(f[x][i] != f[y][i]) {
			x = f[x][i];
			y = f[y][i];
			ans += 1 << (i + 1);
		}
	}
	return ans + 2;
}


int get(int x) {
	return fa[x] == x ? fa[x] : fa[x] = get(fa[x]);
}

void Merge(int x , int y) {
	fa[get(x)] = get(y);
	return;
}

struct Tmp {
	int id , x;
}c[maxN];
bool cmp(Tmp a , Tmp b) {
	return a.x > b.x;
}

int n;

int main() {
//	freopen("c.in","r",stdin);
//	freopen("c.out","w",stdout);
	n = gi();
	for(int i = 1;i < n;++ i) {
		a[i].u = gi();
		a[i].v = gi();
	}
	for(int i = 1;i < n;++ i) {
		add_Node(a[i].u , a[i].v);
		add_Node(a[i].v , a[i].u);
	}
	dfs(1,0);
	for(int i = 1;i <= n;++ i) fa[i] = i;
	for(int i = 1;i <= n;++ i) b[i] = mk(i , i);
	for(int i = 1;i < n;++ i) {
		int x = a[i].u , y = a[i].v;
		int x1 = b[get(x)].fi , y1 = b[get(x)].se;
		int x2 = b[get(y)].fi , y2 = b[get(y)].se;
		Merge(x , y);
		c[1] = (Tmp) {1 , lca(x1 , y1)};
		c[2] = (Tmp) {2 , lca(x2 , y2)};
		c[3] = (Tmp) {3 , lca(x1 , x2)};
		c[4] = (Tmp) {4 , lca(x1 , y2)};
		c[5] = (Tmp) {5 , lca(y1 , x2)};
		c[6] = (Tmp) {6 , lca(y1 , y2)};
		sort(c + 1, c + 7,cmp);
		if(c[1].id == 1) b[get(x)] = mk(x1,y1);
		if(c[1].id == 2) b[get(x)] = mk(x2,y2);
		if(c[1].id == 3) b[get(x)] = mk(x1,x2);
		if(c[1].id == 4) b[get(x)] = mk(x1,y2);
		if(c[1].id == 5) b[get(x)] = mk(y1,x2);
		if(c[1].id == 6) b[get(x)] = mk(y1,y2);
		printf("%d\n",c[1].x);
	}
	return 0;
}

10.18

不知道出题人出一堆模板有什么意义？
而且出题人语文不太好，T2什么拐弯，md，我醉了。
T1：模板题

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}
const int maxN = 200 + 7;

int l1,l2,next[maxN], v[maxN];
char a[maxN][maxN] , b[maxN][maxN];

void get_next(int id){
	int p = 0;next[1] = 0;
	for(int i = 2;i <= l2;++ i){
		while(p > 0 && b[id][i] != b[id][p + 1])p = next[p];
		if(b[id][i] == b[id][p + 1])p ++;
		next[i] = p; 
	}
}
long long Sum = 0;

void kmp(int id2,int id) {
	int p = 0,ans = 0;
	for(int i = 1;i <= l1;++ i){ 
		while(p > 0 && a[id][i] != b[id2][p + 1])p = next[p];
		if(a[id][i] == b[id2][p + 1])p ++;
		if(p == l2) {
			Sum += (i - l2 + 1) * v[id2];
			p = next[p];
		}
	}
	
}

int main(){
	freopen("dream.in","r",stdin);
	freopen("dream.out","w",stdout);
	int n = gi() , m = gi();
	for(int i = 1;i <= m;++ i) scanf("%s",b[i] + 1);
	for(int i = 1;i <= m;++ i) v[i] = gi();
	for(int i = 1;i <= n;++ i) scanf("%s",a[i] + 1);
	
	for(int i = 1;i <= m;++ i) {
		l2 = strlen(b[i] + 1);
		get_next(i);
		for(int j = 1;j <= m;++ j) {
			l1 = strlen(a[j] + 1);
			kmp(i,j);
		}
	}
	cout << Sum;
	return 0;
}

T2:

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define mk make_pair
#define fi first
#define se second
#define int long long

const int INF = -4485090715960753727;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int gx[] = {0,0,1,-1};
const int gy[] = {1,-1,0,0};

const int maxN = 300 + 7;

int f[25][maxN][maxN];
int a[maxN][maxN];

signed main() {
	freopen("corner.in","r",stdin);
	freopen("corner.out","w",stdout);
	int m = gi() , n = gi();
	memset(f , -0x3f,sizeof(f));
	for(int i = 1;i <= n;++ i) {
		for(int j = 1;j <= m;++ j) {
			a[i][j] = gi();
			if(a[i][j] == 0) {
				f[0][i][j] = 0;
			}
		}
	}
	for(int l = 1;l <= 23;++ l) {
		for(int i = 1;i <= n;++ i) {
			for(int j = 1;j <= m;++ j) {
				for(int k = 0;k < 4;++ k) {
					if(a[i][j]) 
					f[l][i][j] = max(f[l][i][j] , f[l - 1][i + gx[k]][j + gy[k]] + a[i][j]);
				}
			}
		}
	}
	int ans = INF;
	for(int i = 1;i <= n;++ i) {
		for(int j = 1;j <= m;++ j) {
			ans = max(ans , f[23][i][j]);
		}
	}
	cout << ans;
	return 0;
}
/*
4 4
1 1 1 1
1 1 0 1
1 1 1 1
1 1 1 1
*/

T3:

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define dbug(x) cout << x << endl
#define mk make_pair
#define fi first
#define se second

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

struct bignum {
	int n;
	int a[3007];
	bignum() {
		n = 0;
		memset(a, 0, sizeof(a));
	}
	bignum(string s) {
		n = s.size();
		memset(a, 0, sizeof(a));
		for (int i = 0; i < n; i++)
			a[i] = s[n - 1 -i] -'0';
	}
	bignum(int s) {
		memset(a, 0, sizeof(a));
		n = 0;
		while (s > 0) {
			a[n] = s % 10;
			s /= 10;
			n++;
		}
	}
	void work() {
		for (int i = 0; i < n; i++) {
			if (a[i] < 0) {
				int tmp = (-a[i] - 1) / 10 + 1;
				a[i] += 10 * tmp;
				a[i + 1] -= tmp;
			}
			if (a[i] >= 10) {
				int tmp = a[i] / 10;
				a[i] -= 10* tmp;
				a[i + 1] += tmp;
				if (i == n - 1 && a[i + 1] > 0) n++;
			}
		}
		while (n > 0 && a[n - 1] == 0) n--;
	}
	void print() {
		for (int i = n - 1; i >= 0; i--)
			cout << a[i];
		cout << endl;
	}
}f[1007],c;

bignum operator + (const bignum &a, const bignum &b) {
	bignum c;
	c.n = max(a.n, b.n);
	for (int i = 0; i < c.n; i++)
		c.a[i] = a.a[i] + b.a[i];
	c.work();
	return c;
}

bignum operator - (const bignum &a, const bignum &b) {
	bignum c;
	c.n = max(a.n, b.n);
	for (int i = 0; i < c.n; i++)
		c.a[i] = a.a[i] - b.a[i];
	c.work();
	return c;
}

bignum operator * (const bignum &a, const bignum &b) {
	bignum c;
	c.n = a.n + b.n - 1;
	for (int i = 0; i < a.n; i++)
		for (int j = 0; j < b.n; j++)
			c.a[i + j] += a.a[i] * b.a[j];
	c.work();
	return c;
}

//int ans[50] = {0,0,1,2,9,44,265,1854,14833,133496,1334961};

int main() {
	freopen("keke.in","r",stdin);
	freopen("keke.out","w",stdout);
	f[1] = 0;f[2] = 1;
	int n = gi();
	for(int i = 3;i <= n;++ i) {
		bignum a(i - 1);
		f[i] = a * (f[i - 1] + f[i - 2]);
	}
	f[n].print();
	return 0;
}

10.16

预计得分：100 + [30,60] + 30 = [160,190]
实际得分：100 + 50 + 0 = 150
？？？？
我T3忘记一次性处理，继而输出答案了，全场只有我自己写了30？？？？（然后挂分了
T1：
刚开始不会做啊，以为是质数不满足，结果cdx大佬开场1min就切了,强。
后来把素数与素数相乘的数先找出来
4(2 * 2)，6(23)，9(33)...blablabla
然后看看能凑成什么数，发现在12之后就有长度为4全部满足了，之后肯定满足。
[1,11]内的手玩，发现只有1,3,5,7,11不满足。
特判掉就可以了。

bool check(int x) {
	if(x == 1 || x == 5 || x == 7 || x == 11 || x == 2 || x == 3) return true;
	return false;
}

signed main() {
	int q = gi();
	while(q --) {check(gi()) ? puts("0") : puts("1");}
	return 0;
}

T2：
我先想了一个\(900 \times n^3 \times q\)的算法，没写，写了一个假n^3DP
然后转瞬间又想到了一个贪心\(900*n*q\)的做法，想到不可能这么简单，就没写。。。。。。(经axm贪心大法怒AT2证明，是正确的....
然后这就接近正解了？？？？
然而我的dp状态不太对吧，233，而且我还比std的二进制运算多了30的常数。
CODE:

#include <bits/stdc++.h>
using namespace std;
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
//#define getchar()(p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
//char buf[1<<21],*p1=buf,*p2=buf;

inline int gi() {
	int x = 0,f = 1;char c = getchar();
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = getchar();}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = getchar();}
	return x * f;
}

int q,n,m,ans,x,s[1005],f[1005];

int main(){
	freopen("b.in","r",stdin);
	freopen("b.out","w",stdout);
	q = gi();
	while(q --) {
		n = gi();m = gi();
		s[0] = 0;
		rep(i , 1, n) {
			x = gi();s[i] = s[i - 1] ^ x;
		}
		ans = 0;
		sep(i , 29, 0) {
			memset(f,0,sizeof(f));
			f[0] = 1;
			for (int j = 1;j <= n;++ j){
				for (int k = 0;k < j;++ k) if(f[k] > 0){
					x = s[j] ^ s[k];
					if ( ((ans & x) >= ans) && (( x & ( 1 << i)) > 0) )
						f[j] = max(f[j] , f[k]+1);
				}
			}
			if(f[n] > m) ans |= (1 << i);
		}
		cout << ans << '\n';
	}
	return 0;
}

T3：
正解不会，这种递推类型的题
状态压缩还是会一点点的。
\(f[S][i]\)表示状态S下价值为i的方案数
转移枚举最后一位是那个数
CODE：

	for(int i = 1;i < (1 << n);++ i) {
		int len = 0;
		for(int j = 1;j <= n;++ j) if(i & (1 << j - 1)) len ++;
		for(int j = 1;j <= n;++ j) {
			if(i & (1 << j - 1)) {
				if(abs(j - len) <= 1)  {
					for(int k = 1;k <= n;++ k) {
						int S1 = i ^ (1 << j - 1);
						f[i][k] += f[S1][k - 1];
						f[i][k] %= p;
					}
				}
				else {
					for(int k = 0;k <= n;++ k) {
						int S1 = i ^ (1 << j - 1);
						f[i][k] += f[S1][k];
						f[i][k] %= p;
					}
				}
			}
		}
	}

10.15

预计得分：70+50+0
实际得分：30+30+0
？？？评测机太慢了吧。。。。
修锅后：70+50+0
T1一直在猜是\(O(n)\)算法，然后想不出，心态炸了。
T2是乱搞多得了20，稍微有点大常数，后面都T了。
仍然不会正解
T3不会。
最后20min才写T1 70，也没心思写我的log算法，想log^2和log差不多，就没写log！
写一下自己的log做法吧，很麻烦就是了：
先不考虑加入修改操作
考虑首先将两个序列合并起来。
然后区间\([l1,r1]\)对应着一段区间，区间\([l2,r2]\)对应着一段区间。
两个区间不相交，区间中间值必定在长度大的区间里。
如果相交，那么相交的部分是可以二分的。
考虑加入修改操作
我们把四个端点分别二分就好啦（常数大警告
反正复杂度是\(log_n\)的（小声BB
正解还是比较妙的，分治。
直接粘solution：

普适性更强的方法。假设当前要取的是区间的第k大，将k折半，放在两个区间的对应位置 s,t上，比较 a[s],b[t]，不妨设a[s]<b[t]，那么答案可以化归至区间[l1,s−1],[l2,r2]的第k2大数（因为a序列比a[s]小的那些数一定可以全部舍去）， 递归即可
T1：

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int,int>
#define int long long
#define mk make_pair
#define fi first 
#define se second

inline int gi() {
	int x = 0, f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}
const int maxN = 5e5 + 7;

int a[maxN], b[maxN];

int kth(int ta[], int sa, int tb[], int sb, int k) {
	if (sa > sb) return kth(tb, sb, ta, sa, k);
	if (sa == 0) return tb[k];
	if (k == 1) return std::min(ta[1], tb[1]);
	int ka = std::min(sa, k/2), kb = k - ka;
	if (ta[ka] < tb[kb]) return kth(ta+ka, sa-ka, tb, sb, k-ka);
	return kth(ta, sa, tb+kb, sb-kb, k-kb);
}

int query(int la, int ra, int lb, int rb) {
	int sa = ra-la+1, sb = rb-lb+1, siz = sa + sb;
	return kth(a+la-1, sa, b+lb-1, sb, siz/2+1);
}

int main() {
	freopen("median.in", "r", stdin);
	freopen("median.out", "w", stdout);
	int n, m;
	n = gi(); m = gi();
	for (int i = 1; i <= n; i++) a[i] = gi();
	for (int i = 1; i <= n; i++) b[i] = gi();
	for (int opt; m--; ) {
		opt = gi();
		if (opt == 2) {
			int la = gi(), ra = gi(), lb = gi(), rb = gi();
			printf("%d\n", query(la, ra, lb, rb));
		} else {
			int p = gi(), pos = gi(), val = gi();
			if (p == 0) a[pos] = val;
			else b[pos] = val;
		}
	}
	return 0;
}

成绩：

10.13

预计得分：100+100+80
实际得分：100+100+40
这场模拟赛有点无聊
开错题了，把T2当成T1了。
随便猜了个结论，开场5min就写完了！
T1忘记概率怎么搞了，直接用方案数算的。。。。。（核心差不多
T3好题，分析性质，一步步优化到正解，由于没有用到核心性质，常数比其他人大很多。（其他人二分用的都是整形，我是实数）
T3比较难写，先咕着
T1

#include <bits/stdc++.h>
using namespace std;
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define gc getchar()
#define int long long

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int maxN = 1000 + 7;

int n , m;
double f1[maxN][maxN * 6];
double f2[maxN][maxN * 6];

void Pre() {
	f2[0][0] = f1[0][0] = 1;
	rep(i , 1, n) {
		rep(j , 0, 6 * i) {
			for(int k = 1;k <= 6;++ k) {
				if(j >= k) 	f1[i][j] += f1[i - 1][j - k];
			}
			f1[i][j] /= 6;
		}
	}
	rep(i , 1, m) {
		rep(j , 0, 6 * i) {
			for(int k = 1;k <= 6;++ k) {
				if(j >= k) 	f2[i][j] += f2[i - 1][j - k];
			}
			f2[i][j] /= 6;
		}
	}
}

void Solve() {
	double ans = 0;
	rep(i , 1, 6 * n) {
		rep(j , 1, i - 1) {
			ans += f1[n][i] * f2[m][j];
		}
	}
	ans *= 100;
	printf("%.2lf%%",ans);
}

signed main() {
	n = gi();m = gi();
	Pre();
	Solve();
	return 0;
}

T2

#include <bits/stdc++.h>
using namespace std;
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define gc getchar()
#define int long long
const int maxN = 1e6 + 7;
const int inf = 0x7fffffff;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

struct Node{
	int a , b;
}a[maxN];
int sum1[maxN] , sum2[maxN];

bool cmp(Node a , Node b) {
	if(a.a == b.a) return a.b < b.b;
	return a.a < b.a;
}

signed main() {
	int n = gi();
	rep(i , 1, n) a[i].a = gi();
	rep(i , 1, n) a[i].b = gi();
	
	sort(a + 1, a + n + 1,cmp);
	
	
	rep(i , 1, n) sum1[i] = a[i].a * a[i].b;
	rep(i , 1, n) sum1[i] += sum1[i - 1];
	rep(i , 1, n) sum2[i] = a[i].b;
	rep(i , 1, n) sum2[i] += sum2[i - 1];
	
	int ans = 0;
	rep(i , 1, n) {
		int sum = 0;
		sum = sum2[i - 1] * a[i].a - sum1[i - 1];
		sum += (sum1[n] - sum1[i]) - a[i].a * (sum2[n] - sum2[i]);
		if(i == 1) ans = sum;
		else ans = min(ans , sum);
	}
	printf("%lld",ans);
	return 0;
}
/*
6
2 1 3 3 2 4
5 1 1 2 2 2
*/

T2得到了正解，

const int maxN = 1e6 + 7;

struct Node {
	int a , b;
}a[maxN];

bool cmp(Node a , Node b) {
	return a.a < b.a;
}

signed main() {
	int n = gi();
	rep(i , 1, n) a[i].a = gi();
	rep(i , 1, n) a[i].b = gi();
	sort(a + 1,a + n + 1,cmp);
	int ans = 0 , sum_b = 0, cur = 0;
	rep(i , 1, n) cur += a[i].b * (a[i].a - a[1].a);
	ans = cur;
	int k1 = 0, k2 = 0;
	rep(i , 2, n) k1 += a[i].b;k2 = a[1].b;
	rep(i , 2, n) {
		cur -= k1 * (a[i].a - a[i - 1].a);
		cur += 1ll * k2 * (a[i].a - a[i - 1].a);
		k1 -= a[i].b;k2 += a[i].b;
		ans = min(ans , cur);
	}
	cout << ans; 
	return 0;
}

T3

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>

using namespace std;

const int MAXN = 1000010;
const int MAXL = 22;

int rmin[MAXL][MAXN], rmax[MAXL][MAXN];

int n, q;
int a[MAXN];

inline void init_rmq() {
    for (int i = 1; i < MAXL; i++)  {
        for (int j = 0; j + (1 << i) <= n; j++) {
            rmin[i][j] = min(rmin[i-1][j], rmin[i-1][j+(1<<(i-1))]);
            rmax[i][j] = max(rmax[i-1][j], rmax[i-1][j+(1<<(i-1))]);
        }
    }
}

inline bool test(int d, int m) {
    int l = 0, i;
    for (i = 0; i < m && l < n; i++) {
        int r = l, smin = 1000000001, smax=0;
        for (int j = MAXL - 1; j >= 0; j--) {
            if (r + (1 << j) <= n) {
                int tmin = rmin[j][r], tmax = rmax[j][r];
                if (max(smax, tmax) - min(smin, tmin) <= d) {
                    smax = max(smax, tmax);
                    smin = min(smin, tmin);
                    r += (1 << j);
                }
            }
        }
        l = r;
    }
    return l == n;
}

inline int work(int m) {
    int l = 0, r = 1000000001;
    while (l < r) {
        int mid = (l + r) >> 1;
        if (test(mid, m)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l;
}

int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%*d%d", &rmin[0][i]);
        rmax[0][i] = rmin[0][i];
    }
    init_rmq();
    scanf("%d", &q);
    for (int i = 0; i < q; i++) {
        int m;
        scanf("%d", &m);
        int ans = work(m);
        if (ans % 2) printf("%d.5\n", ans/2);
        else printf("%d\n", ans/2);
    }
}

10.5

可能是因为坚持一早自习没睡觉，考试特别困。然后一点考试的状态都没有。
连基础的状态压缩都搞不到手了，没开longlong，题意读错导致数组问题等出现了很多问题。
没有合理安排时间，写第一题代码我记得是10点。
预计得分：100 + 22 + 0
实际得分：0+2+0
考完心态爆炸了
仅仅只有T1代码剩下两道题没补

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define int long long
inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int maxN = 3e3 + 7;

int sum[maxN][maxN];
int t1[maxN][maxN];
int t2[maxN][maxN];
int t3[maxN][maxN],t4[maxN][maxN];
int n , q;

void solve() {
	for(int i = 1;i <= n;++ i) {
		for(int j = 1;j <= n;++ j) {
			sum[i][j] = sum[i][j] + sum[i][j - 1] + t1[i][j] + t4[i][j];
		}
	}
	int ans = 0;
	rep(i , 1, n) {
		rep(j , 1, n) {
			ans ^= sum[i][j];
		}
	}
	cout << ans;
}

void work() {
	rep(i , 1, n) {
		rep(j , 1, n) {
			t1[i][j] = t1[i - 1][j] + t2[i][j];
		}
	}
	rep(i , 1, n) {
		rep(j , 1, n) {
			t4[i][j] = t4[i - 1][j - 1] + t3[i][j];
		}
	}
	return ;
}

signed main() {
//	freopen("u.in","r",stdin);
//	freopen("u.out","w",stdout);
	n = gi();q = gi();
	while(q --) {
		int r = gi() , c = gi(),l = gi(),s = gi();
		t2[r][c] += s;
		t2[r + l][c] -= s;
		t3[r][c + 1] -= s;
		t3[r + l][c + l + 1] += s;
 	}
	work();
	solve();
	return 0;
}

10.6

考过的模拟赛，T2是个假题，T3神仙题。
花几分钟写了T1，T3 60，然后就开始看别的课件了。
神仙学弟sjp搜索踩爆状态压缩了，很裸很裸的那种搜索。
牛皮
T2就不补了
T1:

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)

inline int gi() {
	int x = 0, f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
} 
const int maxN = 1e5 + 7;

struct Node {
	int l , x, v;
}a[maxN];

int main() {
	freopen("cruise.in","r",stdin);
	freopen("cruise.out","w",stdout);
	int T = gi();
	while(T --) {
		int n = gi();
		rep(i , 1, n + 1) a[i].l = gi();
		rep(i , 1, n + 1) a[i].x = gi();
		rep(i , 1, n + 1) a[i].v = gi();
//		double t = 1.0 * (a[n + 1].x + a[n + 1].l) / a[n + 1].v;
//		for(int i = n;i >= 2;-- i) {
//			int L = a[i].v * t;
//			if(L >= a[i].x) {
//				t += 1.0 * a[i].l / a[i].v;
//			}else {
//				t = 1.0 * (a[i].x  + a[i].l) / a[i].v;
//			}
//		}
////		printf("%lf\n",t);
//		int L = a[1].v * t;
//		if(L + a[1].l <= a[1].x) {
//				t += 1.0 * a[1].l / a[1].v;
//		}else {
//			t = 1.0 * a[1].x / a[1].v;
//		}
//		printf("%lf",t);
		double ans = 0;
		double sum = 0;
		for(int i = 1;i <= n + 1;i++) {
		  if(i != 1) sum += a[i].l;
		  double t = (a[i].x + sum) / a[i].v;
		  ans = max(t,ans);
		}
		printf("%.10f\n",ans);
	}
	return 0;
}

T3:

#include <bits/stdc++.h>
using namespace std;
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)

const int maxN = 10 + 7;

char a[maxN][maxN];
int f[maxN][maxN][maxN][maxN];

int calc(int x1, int y1, int x2, int y2, int i1, int j1, int i2, int j2) {
	int res = 0;
	for (int i = i1; i <= i2; ++ i) 
		for (int j = j1; j <= j2; ++ j) 
			if (a[i][j] == '#') 
				res += max(max(abs(i-x1), abs(i-x2)), max(abs(j-y1), abs(j-y2)));
	return res;
}

int dp(int x1 , int y1, int x2 , int y2) {
	if(x1 > x2 || y1 > y2) return 0;
	int &res = f[x1][y1][x2][y2];
	if(res > -1) return res;
	res = calc(x1, y1, x2, y2, x1, y1, x1, y2) + dp(x1+1, y1, x2, y2);
	res = min(res, calc(x1, y1, x2, y2, x2, y1, x2, y2) + dp(x1, y1, x2-1, y2));
	res = min(res, calc(x1, y1, x2, y2, x1, y1, x2, y1) + dp(x1, y1+1, x2, y2));
	res = min(res, calc(x1, y1, x2, y2, x1, y2, x2, y2) + dp(x1, y1, x2, y2-1));
	return res;
}

int main() {
	memset(f , -1, sizeof(f));
	
	rep(i , 1, 8) scanf("%s",a[i] + 1);
	
	printf("%d",dp(1,1,8,8));
	return 0;
}

10.11普及模拟赛

普及组的题目，比较简单。
昨天晚上忘记有这个比赛了，最后1h才想起来去看了看题，想了想。
想是想出来了，但是写不动T2，就只写了T1,T2.

T1

给定x,y,a,b（都是在数轴上的点
开始点在x,求到y的最小距离。
有一对传送点（a,b），能从a不花费任何距离到b，或从b不花费任何距离到a.

让x<y,a<b.然后判断走不走(a,b)哪个更优

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)

inline int gi() {
	int x = 0, f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

int main() {
	int x = gi() , y = gi(), a = gi(), b = gi();
	if(x > y) swap(x , y);
	if(a > b) swap(a , b);
	int ans = y - x;
	ans = min(ans , abs(x - a) + abs(y - b));
	printf("%d",ans);
	return 0;
}

T2

很长的模拟题面

模拟就行了，额

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int , int>
#define mk make_pair
#define fi first
#define se second
#define CL(x) memset(x , 0, sizeof(x));
#define int long long
const int maxN = 1e7 + 7;

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

int a[10];
int b[10] , c[10], d[10];

signed main() {
	int n;
	while(scanf("%lld",&n) && n) {
		int q = sqrt(n * 10);
		CL(b);
		int tmp = n;
		while(tmp) {b[tmp % 10] ++;tmp /= 10;}
		if(b[0] > 0) {
			printf("0 * 0 = 0\n");
		}
		for(int i = 1;i < q;++ i) {
			int tmp = i;
			CL(a);
			while(tmp) a[tmp % 10] ++ , tmp /= 10;
			bool flag = true;
			rep(j , 0, 9) {
				if(a[j] > b[j]) flag = false;
			}
			if(flag) {
				int x = i * i;
				CL(d);
				while(x) {
					d[x % 10] ++;
					x /= 10;
				}
				bool flag = true;
				rep(j , 0, 9) if(d[j] > b[j]) flag = false;
				if(flag) printf("%lld * %lld = %lld\n",i,i,1ll * i * i);
			}
		}
	}
	return 0;
}

T3

给定一个背包容量m和很多物品，求多少方案使得\(m - min(不选的物品) < 选择物品之和\)

从小到大排序，枚举最小不选的物品是哪一个，然后先把前面的都选了，然后背包求方案数
注意全都选择的情况。

#include <bits/stdc++.h>
using namespace std;
#define gc getchar()
#define rep(i , x, y) for(int i = x;i <= y;++ i)
#define sep(i , x, y) for(int i = x;i >= y;-- i)
#define PII pair<int , int>
#define mk make_pair
#define fi first
#define se second
#define CL(x) memset(x , 0, sizeof(x));

inline int gi() {
	int x = 0,f = 1;char c = gc;
	while(c < '0' || c > '9') {if(c == '-')f = -1;c = gc;}
	while(c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc;}
	return x * f;
}

const int maxN = 30 + 7;
const int maxM = 1000 + 7;

int n , m, a[maxN], f[maxN][maxM];

bool cmp(int a , int b) {return a > b;}

void Read() {
	n = gi();m = gi();
	int sum = 0;
	rep(i , 1, n) a[i] = gi();
	sort(a + 1, a + n + 1);
}

void Solve() {
	int sum = 0, ans = 0;
	rep(i , 1, n) sum += a[i];
	if(sum <= m){puts("1");return;} 
	rep(i , 1, n) {
		sum = 0;
		CL(f);
		rep(j , 1, i - 1) sum += a[j];
		if(sum > m) break;
		f[i][sum] = 1;
		rep(k , i + 1, n) {
			rep(j , 0, m) {
				f[k][j] = f[k - 1][j];
				if(j >= a[k]) f[k][j] += f[k - 1][j - a[k]];
			}
		}
		rep(j , m - a[i] + 1, m) {
			ans += f[n][j];
		}
	}
	printf("%d\n",ans);
	return ;
}

int main() {
	int T = gi();
	while(T --) {
		Read();
		Solve();
	}
	return 0;
}

T4:
比较容易想到的套路题吧。
首先离散化，然后按照询问的大小从大到小排列，直接并查集合并。