CSP认证

202203-1 未初始化警告

模拟

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, k;
	cin >> n >> k;
	vector<int> ok(n + 1, 0);
	ok[0] = 1;
	int ans = 0;
	for(int i = 0; i < k; ++ i) {
		int x, y;
		cin >> x >> y;
		if(ok[y] == 0) ans ++;
		ok[x] = 1;
	}
	cout << ans << "\n";
	return 0;
}

202203-2 出行计划

\(t - c + 1 <= q + k <= t\), 偏移一下，做差分前缀和

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;

int sum[500010];
int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, m, k;
	cin >> n >> m >> k;
	int dlt = 200000;
	for(int i = 0; i < n; ++ i) {
		int t, c;
		cin >> t >> c;
		sum[t - c + 1 + dlt] += 1;
		sum[t + 1 + dlt] -= 1;
	}
	for(int i = 0; i < 500010; ++ i) sum[i] += sum[i - 1];
	for(int i = 0, x; i < m; ++ i) {
		cin >> x;
		cout << sum[x + k + dlt] << "\n";
	}
	return 0;
}

202203-3 计算资源调度器

按题意模拟，写的巨丑无比

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;

int n, m, k, l[N];
int f, a, na, pa, paa, paar;
int cnt[N];
map<int, vector<int>> region;
map<int, vector<int>> point;

vector<int> Filter(int a, int na, int pa, int paa, int paar) {
	vector<int> ans(n);
	for(int i = 0; i < n; ++ i) ans[i] = i + 1;
	if(na) {
		vector<int> res;
		for(int &x : ans) 
			if(l[x] == na) 
				res.push_back(x);
		ans = res;
	}
	if(pa) {
		vector<int> res;
		map<int, int> t;
		for(int &x : region[pa]) {
			t[x] = 1;
		}
		for(int &x : ans) 
			if(t.count(l[x])) {
				res.push_back(x);
			}
		ans = res;
	}
	if(paa) {
		vector<int> res;
		map<int, int> t;
		for(int &x : point[paa]) {
			t[x] = 1;
		}
		for(int &x : ans) 
			if(!t.count(x)) {
				res.push_back(x);
			}
		if(paar == 0 and res.size() == 0) return ans;
		ans = res;
	}
	return ans;
}

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	cin >> n >> m;
	for(int i = 1; i <= n; ++ i) cin >> l[i];
	cin >> k;
	for(int i = 1; i <= k; ++ i) {
		cin >> f >> a >> na >> pa >> paa >> paar;
		for(int j = 1; j <= f; ++ j) {
			vector<int> ans = Filter(a, na, pa, paa, paar);
			if(ans.size() == 0) {
				cout << "0" << " \n"[j == f];
				continue;
			}
			sort(ans.begin(), ans.end(), [&](const int &x, const int &y) {
				if(cnt[x] != cnt[y])
					return cnt[x] < cnt[y];
				return x < y;
			});
			cout << ans[0] << " \n"[j == f];
			cnt[ans[0]] ++;
			region[a].push_back(l[ans[0]]);
			point[a].push_back(ans[0]);
		}
	}
	return 0;
}

202203-4 通信系统管理

set或者可删除堆维护一下，维护“通信对”细节较多

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 100010;
struct Node {
	LL w;
	int v;
	bool operator < (const Node& t) const {
		if(w != t.w) return w < t.w;
		return v > t.v;
	}
};
struct Query {
	int u, v, w;
};
int n, m, cnt1, cnt2;
vector<Query> del[N];
unordered_map<LL, LL> mp;
unordered_map<int, int> pp;
set<Node> S[N];

void _add(int u, int v, int w) {
	LL st = 1LL * (u - 1) * n + v;
	int L = S[u].size();
	S[u].erase({mp[st], v});
	mp[st] += w;
	if(mp[st] != 0)
		S[u].insert({mp[st], v});
	int R = S[u].size();
	if(L && !R) cnt1 ++; 
	if(!L && R) cnt1 --;
	if(S[u].size() == 0) {
		if(pp.count(pp[u]) && pp[pp[u]] == u) cnt2 --;
		pp.erase(u);
	}
	if(S[u].size() && pp[u] != S[u].rbegin()->v) {
		if(pp.count(pp[u]) && pp[pp[u]] == u) cnt2 --;
		pp.erase(u);
		int to = S[u].rbegin()->v;
		if(S[to].size() && S[to].rbegin()->v == u) {
			pp[to] = u;
			pp[u] = to;
			cnt2 ++;
		}
	}
	if(S[u].size() && S[v].size() && S[u].rbegin()->v == v and S[v].rbegin()->v == u)
		if(pp[u] != v) pp[u] = v, pp[v] = u, cnt2 ++; 
}

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	cin >> n >> m;
	cnt1 = n;
	for(int i = 0; i < m; ++ i) {
		int k; cin >> k;
		for(auto &q: del[i]) _add(q.u, q.v, q.w), _add(q.v, q.u, q.w);
		for(int j = 0; j < k; ++ j) {
			int u, v, x, y; cin >> u >> v >> x >> y;
			_add(u, v, x), _add(v, u, x);
			if(i + y < m) del[i + y].push_back({u, v, -x});
		}
		int l; cin >> l;
		for(int j = 0, x; j < l; ++ j) {
			cin >> x;
			if(S[x].size() == 0) cout << "0" << "\n";
			else cout << (S[x].rbegin()->v) << "\n";
		}
		int p, q; cin >> p >> q;
		if(p) cout << cnt1 << "\n";
		if(q) cout << cnt2 << "\n";
	}
	return 0;
}

202112-1 序列查询

模拟

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
using LL = long long;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, N;
	cin >> n >> N;
	vector<int> A(n + 2);
	for(int i = 1; i <= n; ++ i) cin >> A[i];
	A[n + 1] = N;
	LL ans = 0;
	for(int i = 0; i <= n; ++ i) 
		ans += 1LL * (A[i + 1] - A[i]) * i;
	cout << ans << "\n";
	return 0;
}

202112-2 序列查询新解

搞成两组区间，在重叠的地方计算

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
using LL = long long;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, N;
	cin >> n >> N;
	vector<int> A(n + 2), B;
	for(int i = 1; i <= n; ++ i) cin >> A[i];
	A[n + 1] = N;
	int t = 0;
	while(t <= N) {
		B.push_back(t);
		t += N / (n + 1);
	}
	if(B.back() != N) B.push_back(N);
	LL ans = 0;
	for(int i = 0, j = 0; i <= n; ++ i) {
		while(j + 1 < B.size() && B[j + 1] < A[i + 1]) {
			ans += 1LL * abs(j - i) * (B[j + 1] - max(A[i], B[j]));
			j ++;
		}
		ans += 1LL * abs(j - i) * (A[i + 1] - max(A[i], B[j]));	
	}
	cout << ans << "\n";
	return 0;
}

202112-3 登机牌条码

前几个点按题意模拟，后半部分模拟多项式除法、乘法，过程中取模

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
const int P = 929;

int _type(char &x) {
	if(x >= '0' && x <= '9') return 0;
	if(x >= 'A' && x <= 'Z') return 1;
	if(x >= 'a' && x <= 'z') return 2;
}

int _val(char &x) {
	if(_type(x) == 0) return x - '0';
	if(_type(x) == 1) return x - 'A';
	if(_type(x) == 2) return x - 'a';
}

int _tran(char &x, char &y) {
	int tx = _type(x), ty = _type(y);
	if(tx == ty) return 0;
	if(tx == 0 && ty == 1) return 28;
	if(tx == 2 && ty == 0) return 28;
	if(tx == 1 && ty == 0) return 28;
	if(tx == 1 && ty == 2) return 27;
	if(tx == 0 && ty == 2) return 27;
	if(tx == 2 && ty == 1) return 56;
}

using poly = vector<int>;

int power(int a, int b) {
	int res = 1;
	for(; b; b >>= 1, a = a * a % P) 
		if(b & 1) res = res * a % P;
	return res;
}

poly _mul(poly &a, poly &b) {
	poly c(a.size() + b.size() - 1);
	for(int i = 0; i < (int)a.size(); ++ i)
		for(int j = 0; j < (int)b.size(); ++ j)
			c[i + j] = (c[i + j] + a[i] * b[j] % P) % P;
	return c;
}

poly _div(poly &a, poly &b, int k) {
	poly c(k);
	int na = a.size(), nb = b.size();
	for(int i = na - 1; i >= 0; -- i) {
		int bs = i - nb + 1;
		if(bs < 0) break;
		int xs = a[i] * power(b.back(), P - 2) % P;
		for(int j = 0; j < nb; ++ j) 
			a[j + bs] = (a[j + bs] - xs * b[j] % P + P) % P;
	}
	for(int i = 0; i < k; ++ i) c[i] = a[i];
	while(c.back() == 0) c.pop_back();
	return c;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr); cin.tie(nullptr);
   	int w, s;
   	string str;
   	cin >> w >> s >> str;
   	vector<int> v;
   	char START = 'A';
   	for(int i = 0; i < (int)str.size(); ++ i) {	
   		int tr;
   		if(i) tr = _tran(str[i - 1], str[i]);
   		else tr = _tran(START, str[i]);
   		if(tr) {
   			if(tr == 56) v.push_back(28), v.push_back(28);
   			else v.push_back(tr);
   		}
   		v.push_back(_val(str[i]));
   	}
   	if(v.size() & 1) v.push_back(29);
   	vector<int> ans;
   	int k = 0;
	if(s != -1) k = (1 << (s + 1));
   	ans.push_back((v.size() / 2 + w - 1) / w * w);
   	for(int i = 0; i < (int)v.size(); i += 2) ans.push_back(v[i] * 30 + v[i + 1]);
	while((ans.size() + k) % w) ans.push_back(900);
	ans[0] = ans.size();
	for(int &x : ans) cout << x << "\n";
	if(s != -1) {
		poly g(1, 1);
		for(int i = 1, th = 1; i <= k; ++ i) {
			th = th * 3 % P;
			poly t = {-th, 1};
			g = _mul(g, t);
		}
		poly d(ans.size() + k);
		for(int i = 0; i < ans.size(); ++ i) d[ans.size() - 1 - i + k] = ans[i];
		poly r = _div(d, g, k);
		reverse(r.begin(), r.end());
		for(int &x : r) cout << (-x % P + P) % P << "\n";
	}
    return 0;
}

202112-4 磁盘文件操作

离散化的时候记录两侧的点，线段树维护权值、编号、状态即可，不难写

Sample Code (C++)

#include <bits/stdc++.h>
#define tm (tl + tr >> 1)
#define ls (u << 1)
#define rs (ls | 1)
#define FF 0
#define OO 1
#define RR 2
using namespace std;
using PII = pair<int, int>;
const int N = 1000010;
struct Query {
	int op, id, l, r, x, p;
};
int n, m, k;
int w[N << 2], idx[N << 2], sta[N << 2];

void pushup(int u) {
	w[u] = (w[ls] == w[rs] ? w[ls] : INT_MAX);
	idx[u] = (idx[ls] == idx[rs] ? idx[ls] : INT_MAX);
	sta[u] = (sta[ls] == sta[rs] ? sta[ls] : INT_MAX);
}

void pushdown(int u) {
    if(w[u] != INT_MAX) w[ls] = w[rs] = w[u];
	if(idx[u] != INT_MAX) idx[ls] = idx[rs] = idx[u];
    if(sta[u] != INT_MAX) sta[ls] = sta[rs] = sta[u];
}

void build(int u, int tl, int tr) {
	if(tl == tr) return;
	build(ls, tl, tm);
	build(rs, tm + 1, tr);
	pushup(u);
}

int query_pos(int u, int tl, int tr, int l, int id) {
    if(sta[u] == FF || sta[u] == RR || idx[u] == id) return tr;
    if(sta[u] == OO && idx[u] != INT_MAX) return -1; 
    pushdown(u);
    if(l <= tm) {
        int L = query_pos(ls, tl, tm, l, id);
    	if(L < tm) return L;
        int R = query_pos(rs, tm + 1, tr, l, id);
        return R == -1 ? L : R;
    }
    return query_pos(rs, tm + 1, tr, l, id);
};

void write_val(int u, int tl, int tr, int l, int r, int id, int x) {
	if(tl >= l and tr <= r) { 
        sta[u] = OO;
        idx[u] = id;
        w[u] = x;
        return;
    }
    pushdown(u);
    if(l <= tm) write_val(ls, tl, tm, l, r, id, x);
    if(r > tm) write_val(rs, tm + 1, tr, l, r, id, x);
    pushup(u);
}

bool query_del(int u, int tl, int tr, int l, int r, int id) {
	if(tl >= l and tr <= r) { 
        if(sta[u] == OO && idx[u] == id) return true;
        else return false;
    }
    pushdown(u);
    bool res = true;
    if(l <= tm) res &= query_del(ls, tl, tm, l, r, id);
    if(r > tm) res &= query_del(rs, tm + 1, tr, l, r, id);
    return res;
}

bool query_rec(int u, int tl, int tr, int l, int r, int id) {
	if(tl >= l and tr <= r) {
        if(sta[u] == RR && idx[u] == id) return true;
        else return false;
    }
    pushdown(u);
    bool res = true;
    if(l <= tm) res &= query_rec(ls, tl, tm, l, r, id);
    if(r > tm) res &= query_rec(rs, tm + 1, tr, l, r, id);
    return res;
}

void del(int u, int tl, int tr, int l, int r, int id) {
	if(tl >= l and tr <= r) {
        sta[u] = RR;
        return;
    }
    pushdown(u);
    if(l <= tm) del(ls, tl, tm, l, r, id);
    if(r > tm) del(rs, tm + 1, tr, l, r, id);
    pushup(u);
}

void rec(int u, int tl, int tr, int l, int r, int id) {
	if(tl >= l and tr <= r) {
        sta[u] = OO;
        return;
    }
    pushdown(u);
    if(l <= tm) rec(ls, tl, tm, l, r, id);
    if(r > tm) rec(rs, tm + 1, tr, l, r, id);
    pushup(u);
}

pair<int, PII> query_ans(int u, int tl, int tr, int p) {
	if(tl == p and tr == p) return {sta[u], {idx[u], w[u]}};
	pushdown(u);
	if(p <= tm) return query_ans(ls, tl, tm, p);
	else return query_ans(rs, tm + 1, tr, p);
}

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	cin >> n >> m >> k;
	vector<Query> Q(k);
	vector<int> nums;
	for(int i = 0; i < k; ++ i) {
		int op = 0, id = 0, l = 0, r = 0, x = 0, p = 0;
		cin >> op;
		if(op == 0) cin >> id >> l >> r >> x;
		if(op == 1 || op == 2) cin >> id >> l >> r;
		if(op == 3) cin >> p;
		Q[i] = {op, id, l, r, x, p};
		if(op != 3) {
			nums.push_back(l);
			nums.push_back(r);
			if(l != 1) nums.push_back(l - 1);
			if(r != m) nums.push_back(r + 1);
		}
		else nums.push_back(p);
	}
	nums.push_back(0);
	sort(nums.begin(), nums.end());
	nums.erase(unique(nums.begin(), nums.end()), nums.end());
	m = nums.size();
	for(int i = 0; i < k; ++ i) {
		if(Q[i].op != 3) {
			Q[i].l = lower_bound(nums.begin(), nums.end(), Q[i].l) - nums.begin();
			Q[i].r = lower_bound(nums.begin(), nums.end(), Q[i].r) - nums.begin();
		}
		else Q[i].p = lower_bound(nums.begin(), nums.end(), Q[i].p) - nums.begin();
	}
	build(1, 1, m - 1);
	for(int i = 0; i < k; ++ i) {
		if(Q[i].op == 0) {
			Q[i].r = min(Q[i].r, query_pos(1, 1, m - 1, Q[i].l, Q[i].id));
			if(Q[i].r != -1) write_val(1, 1, m - 1, Q[i].l, Q[i].r, Q[i].id, Q[i].x);
			cout << (Q[i].r == -1 ? -1 : nums[Q[i].r]) << "\n";
		}
		if(Q[i].op == 1) {
			if(query_del(1, 1, m - 1, Q[i].l, Q[i].r, Q[i].id)) {
				del(1, 1, m - 1, Q[i].l, Q[i].r, Q[i].id);
				cout << "OK\n";
			}
			else cout << "FAIL\n";
		}
		if(Q[i].op == 2) {
			if(query_rec(1, 1, m - 1, Q[i].l, Q[i].r, Q[i].id)) {
				rec(1, 1, m - 1, Q[i].l, Q[i].r, Q[i].id);
				cout << "OK\n";
			}
			else cout << "FAIL\n";
		} 
		if(Q[i].op == 3) {
			pair<int, PII> ans = query_ans(1, 1, m - 1, Q[i].p);
			if(ans.first == OO) cout << ans.second.first << " " << ans.second.second << "\n";
			else cout << "0 0\n";
		}
	}
	return 0;
}

202109-1 数组推导

\(mx = \sum B_i\)
\(mn = \sum \begin{cases}0&B_i=B_{i - 1}\\B_i&B_i > B_{i -1}\end{cases}\)

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n;
	cin >> n;
	int mx = 0, mn = 0, lst = -1;
	for(int i = 0, x; i < n; ++ i) {
		cin >> x;
		mx += x;
		if(x > lst) mn += x;
		lst = x;
	}
	cout << mx << "\n" << mn << "\n";
}

202109-2 非零段划分

记一下每个数出现的位置，顺着算一遍

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
const int N = 10010;
vector<int> p[N];

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n;
	cin >> n;
	for(int i = 0, x; i < n; ++ i) {
		cin >> x; 
		p[x].push_back(i);
	}
	vector<int> st(n, 0);
	int cnt = 1, mx = 0;
	for(int i = 0; i < N; ++ i) { 
		for(int &x : p[i]) {
			if(x == 0) cnt -= st[x + 1];
			else if(x == n - 1) cnt -= st[x - 1]; 
			else {
				cnt += (st[x - 1] + st[x + 1] == 0);
				cnt -= (st[x - 1] + st[x + 1] == 2);
			}
			st[x] = 1;
		}
		mx = max(mx, cnt);
	}
	cout << mx << "\n";
	return 0;
}

202104-1 灰度直方图

模拟

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, m, L;
	cin >> n >> m >> L;
	vector<int> h(L);
	for(int i = 0; i < n; ++ i) 
		for(int j = 0; j < m; ++ j) {
			int x; 
			cin >> x;
			h[x] ++;
		}
	for(int i = 0; i < L; ++ i) cout << h[i] << " \n"[i == L - 1];
	return 0;
}

202104-2 邻域均值

前缀和

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	int n, L, r, t;
	cin >> n >> L >> r >> t;
	vector<vector<int>> A(n + 1, vector<int>(n + 1));
	for(int i = 1; i <= n; ++ i)
		for(int j = 1; j <= n; ++ j)
			cin >> A[i][j];
	for(int i = 1; i <= n; ++ i)
		for(int j = 1; j <= n; ++ j)
			A[i][j] += A[i - 1][j] + A[i][j - 1] - A[i - 1][j - 1];
	int ans = 0;
	for(int i = 1; i <= n; ++ i)
		for(int j = 1; j <= n; ++ j) {
			int x1 = max(1, i - r), x2 = min(n, i + r);
			int y1 = max(1, j - r), y2 = min(n, j + r);
			int sum = A[x2][y2] - A[x1 - 1][y2] - A[x2][y1 - 1] + A[x1 - 1][y1 - 1];
			ans += (sum <= t * (x2 - x1 + 1) * (y2 - y1 + 1));
		}
	cout << ans << "\n";
	return 0;
}

202104-3 DHCP服务器

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
const int N = 10010;

int n, m, t_def, t_max, t_min;
string h;
struct IP {
	int state; // [0, 4) 
	int t;
	string owner;
}ip[N];

void update_ips_state(int tc) {
	for(int i = 1; i <= n; ++ i)
		if(ip[i].t && ip[i].t <= tc) {
			if(ip[i].state == 1) {
				ip[i].state = 0;
				ip[i].owner = "";
				ip[i].t = 0;
			}
			else {
				ip[i].state = 3;
				ip[i].t = 0;
			}
		}
}

int get_ip_by_state(int state) {
	for(int i = 1; i <= n; ++ i)
		if(ip[i].state == state)
			return i;
	return 0;
}

int get_ip_by_owner(string client) {
	for(int i = 1; i <= n; ++ i)
		if(ip[i].owner == client)
			return i;
	return 0;
}

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); 
	cin >> n >> t_def >> t_max >> t_min >> h >> m;
	while(m -- ) {
		int tc;
		string client, server, type;
		int id, te;
		cin >> tc >> client >> server >> type >> id >> te;
		if(server != h && server != "*" && type != "REQ") continue;
		if(type != "DIS" && type != "REQ") continue;
		if(server == "*" && type != "DIS" || server == h && type == "DIS") continue;
		update_ips_state(tc);
		if(type == "DIS") {
			int k = get_ip_by_owner(client);
			if(!k) k = get_ip_by_state(0);
			if(!k) k = get_ip_by_state(3);
			if(!k) continue;
			ip[k].state = 1, ip[k].owner = client;
			if(!te) ip[k].t = tc + t_def;
			else {
				int t = te - tc;
				t = max(t, t_min), t = min(t, t_max);
				ip[k].t = tc + t;
			}
			cout << h << ' ' << client << ' ' << "OFR" << ' ' << k << ' ' << ip[k].t << "\n";
		}
		else {
			if(server != h) {
				for(int i = 1; i <= n; ++ i)
					if(ip[i].owner == client && ip[i].state == 1) {
						ip[i].state = 0;
						ip[i].owner = "";
						ip[i].t = 0;
					}
					continue;
			}
			if(!(id >= 1 && id <= n && ip[id].owner == client))
				cout << h << ' ' << client << ' ' << "NAK" << ' ' << id << ' ' << 0 << "\n";
			else {
				ip[id].state = 2;
				if(!te) ip[id].t = tc + t_def;
				else {
					int t = te - tc;
					t = max(t, t_min), t = min(t, t_max);
					ip[id].t = tc + t;
				}
				cout << h << ' ' << client << ' ' << "ACK" << ' ' << id << ' ' << ip[id].t << "\n";
			}
		}
	}
	return 0;
}

202104-4 校门外的树

Sample Code (C++)

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
const int N = 1010, M = 100010, P = 1e9 + 7;

int n, a[N], dp[N];
vector<int> q[M];
bool st[M];

int main() {
	ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
	for(int i = 1; i < M; ++ i)
		for(int j = i * 2; j < M; j += i)
			q[j].push_back(i);
	cin >> n;
	for(int i = 0; i < n; ++ i) cin >> a[i];
	dp[0] = 1;
	for(int i = 1; i < n; ++ i) {
		memset(st, 0, sizeof st);
		for(int j = i - 1; j >= 0; -- j) {
			int d = a[i] - a[j], cnt = 0;
			for(int &k : q[d])
				if(!st[k]) {
					cnt ++;
					st[k] = 1;
				}
			st[d] = 1;
			dp[i] = (dp[i] + 1LL * dp[j] * cnt % P) % P;
		}
	}		
	cout << dp[n - 1] << "\n";
	return 0;
}