CCPC－Wannafly 夏季欢乐赛题解

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#CCPC-Wannafly

本文解析了多项算法竞赛题目，涵盖完全k叉树、距离产生美、烤面包片等，深入探讨解题思路与代码实现，是算法学习者的宝贵资源。

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博主又复活了（因为文化课作业太多写不完了所以继续自爆自弃

A.完全k叉树

签到题，考虑最底层的点的距离即可，注意细节（成功拉低了平均通过率）

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

int main() {
	int T = read();
	while (T--) {
	LL k = read(), n = read(), x = 1, now = 1, cnt = 0;
	if (k == 1) {
		printf("%d\n", n - 1);
		continue;
	}
	while (x < n) {
		cnt++;
		now *= k;
		x += now;
	}
	x -= now;
	cnt--;
	LL ans = 0, lft = n - x;
	if (lft <= now / k) ans = max(cnt * 2 + 1, lft == 1 ? 1LL : 2LL);
	else ans = cnt * 2 + 2;
	printf("%d\n", ans);
	}
	return 0;
}

B.距离产生美

签到题，每次当且仅当 $a_i-a_{i-1}|<k$ 时，把 $a_i$ 改成 $i n f$

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 100010;
	
int a[N], used[N];
inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

int main() {
	int n = read(), k = read(), ans = 0;
	For(i, 1, n) a[i] = read();
	For(i, 2, n) {
		if (abs(a[i] - a[i - 1]) < k && !used[i - 1]) {
			ans++;
			used[i] = 1;
		}
	}
	printf("%d\n", ans);
	return 0;
}

C.烤面包片

签到题， $mod≤1e9,4!!>1e9mod\leq 1e9,4!!>1e9$ 所以只需考虑 $n≤3n\leq 3$ 的情况，其他都输出 $0$ ，注意 $0!!! = 1$

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

inline int mul(int x, int y, int M) {return (int)(1LL * x * y % M);}
inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

int main() {
	int n = read(), M = read();
	if (n == 0 || n == 1) printf("%d\n", 1 % M);
	else if (n == 2) printf("%d\n", 2 % M);
	else if (n == 3) {
		int now = 1;
		For(i, 1, 720) now = mul(now, i, M);
		printf("%d\n", now);
	}
	else puts("0");
		
	return 0;
}

D.茶颜悦色

普通题，将y坐标离散对x这一维线性枚举长度为k的所有点，对y这一维用扫描线维护只对[y-k,y]有贡献，时间复杂度 $O(nlog⁡n)\mathcal{O(n\log n)}$ （~~给茶颜悦色打广告可还行~~

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int N = 100010;
struct Point {
	int x, y;
}p[N];
struct Node {
	Node *ls, *rs;
	int tag, v;
	Node() {
		ls = rs = 0;
		tag = v = 0;
	}
}*root;
int py[N << 1], cnty;

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

inline bool cmp(Point a, Point b) {
	return a.x < b.x || a.x == b.x && a.y < b.y;
}

inline void pushdown(Node *&rt) {
	if (!rt -> ls) rt -> ls = new Node();
	if (!rt -> rs) rt -> rs = new Node();
	rt -> ls -> v += rt -> tag;
	rt -> rs -> v += rt -> tag;
	rt -> ls -> tag += rt -> tag;
	rt -> rs -> tag += rt -> tag;
	rt -> tag = 0;
}

inline void pushup(Node *&rt) {
	rt -> v = max(rt -> ls -> v, rt -> rs -> v);
}

inline void modify(int l, int r, int L, int R, int val, Node *&rt) {
	if (!rt) rt = new Node();
	if (r < L || R < l) return;
	if (L <= l && r <= R) {
		rt -> tag += val;
		rt -> v += val;
		return;
	}
	int mid = l + r >> 1;
	pushdown(rt);
	modify(l, mid, L, R, val, rt -> ls); 
	modify(mid + 1, r, L, R, val, rt -> rs);
	pushup(rt);
}

int main() {
	int n = read(), k = read();
	py[++cnty] = 1;
	For(i, 1, n) {
		p[i].x = read(), p[i].y = read();
		py[++cnty] = p[i].y;
		py[++cnty] = p[i].y - k;
	}
	sort(py + 1, py + cnty + 1);
	int ny = unique(py + 1, py + cnty + 1) - py - 1;
	sort(p + 1, p + n + 1, cmp);
	int l = 1, r = 1;
	for(; r <= n; ++r) {
		if (p[r].x - p[l].x > k) break;
		modify(1, ny, lower_bound(py + 1, py + ny + 1, p[r].y - k) - py, lower_bound(py + 1, py + ny + 1, p[r].y) - py, 1, root);
	}
	int ans = 0;
	for(; r <= n; ++r) {
		ans = max(ans, root -> v);
		for(; p[r].x - p[l].x > k; ++l) modify(1, ny, lower_bound(py + 1, py + ny + 1, p[l].y - k) - py, lower_bound(py + 1, py + ny + 1, p[l].y) - py, -1, root);
		modify(1, ny, lower_bound(py + 1, py + ny + 1, p[r].y - k) - py, lower_bound(py + 1, py + ny + 1, p[r].y) - py, 1, root);
	}
	ans = max(ans, root -> v);
	printf("%d\n", ans);
	return 0;
}

E.飞行棋

普通题，考虑在 $[d - k, d - 1]$ 中的每一个位置都有 $1k\frac{1}{k}$ 的概率到达终点，还有 $k−1k\frac{k-1}{k}$ 的概率到达另一个 $[d - k, d - 1]$ 的点，所以期望步数是 $k$ 。从终点出发逆向dp， $dp[x]=k(x∈[1,k]),dp[x]=1+1k∑i=x−kx−1dp[i]dp[x]=k(x\in [1,k]),dp[x]=1+\frac{1}{k}\sum_{i=x-k}^{x-1}dp[i]$ ，由于d较大，使用矩阵加速即可。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 25;

int K, inv;
inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

inline void upd(int &x, int y) {x += y; x -= x >= MOD ? MOD : 0;}
inline int dec(int x, int y) {x -= y; x += x < 0 ? MOD : 0; return x;}
inline int add(int x, int y) {x += y; x -= x >= MOD ? MOD : 0; return x;}
inline int mul(int x, int y) {return 1LL * x * y % MOD;};

inline int qui_pow(int x, int y) {
	int ret = 1;
	for(; y; y >>= 1) {
		if (y & 1) ret = mul(ret, x);
		x = mul(x, x);
	}
	return ret;
}

struct Matrix {
	int num[N][N];
	Matrix operator * (const Matrix &A) const {
		Matrix ret;
		memset(ret.num, 0, sizeof ret.num);
		For(i, 1, K) For(j, 1, K) For(k, 1, K) upd(ret.num[i][j], mul(num[i][k], A.num[k][j]));
		return ret;
	}
}init, cha;

inline Matrix mat_pow(Matrix x, LL y) {
	Matrix ret;
	memset(ret.num, 0, sizeof ret.num);
	For(i, 1, K) ret.num[i][i] = 1;
	for(; y; y >>= 1) {
		if (y & 1) ret = ret * x;
		x = x * x;
	}
	return ret;
}

int main() {
	LL d = read();
	K = read(), inv = qui_pow(K, MOD - 2);
	For(i, 1, K) init.num[1][i] = K, cha.num[i][1] = inv;
	init.num[1][K + 1] = cha.num[K + 1][1] = cha.num[K + 1][K + 1] = 1;
	For(i, 2, K) cha.num[i - 1][i] = 1;
	K++;
	init = init * mat_pow(cha, d - K + 1);
	printf("%d\n", init.num[1][1]);
	return 0;
}

F.三元组

普通题，假设 $ai+aj≤bi+bja_i+a_j\leq b_i+b_j$ ，所以 $2(ai+aj)≤bi+bj,2ai−bi+2aj−bj≤02(a_i+a_j)\leq b_i +b_j,2a_i-b_i+2a_j-b_j\leq 0$ ，只需按 $2a_i-b_i$ 排序即可， $ai+aj≥bi+bja_i+a_j\geq b_i+b_j$ 亦然

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 100010;

inline void upd(int &x, int y) {x += y; x -= x >= MOD ? MOD : 0;}
inline int dec(int x, int y) {x -= y; x += x < 0 ? MOD : 0; return x;}
inline int add(int x, int y) {x += y; x -= x >= MOD ? MOD : 0; return x;}
inline int mul(int x, int y) {return 1LL * x * y % MOD;};

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

struct Node {
	int a, b, c, v;
}p[N];
int v[N], sum[N], n;

inline bool cmp(Node x, Node y) {return x.v < y.v;}
 
inline int solve() {
	int ret = 0;
	For(i, 1, n) p[i].v = 2 * p[i].a - p[i].b;
	sort(p + 1, p + n + 1, cmp);
	sum[0] = v[0] = 0;
	For(i, 1, n) {
		v[i] = p[i].v;
		sum[i] = add(sum[i - 1], p[i].c);
		upd(ret, mul(sum[upper_bound(v + 1, v + i + 1, -v[i]) - v - 1], p[i].c));
	}
	return ret;
}
		
int main() {
	n = read();
	For(i, 1, n) p[i].a = read(), p[i].b = read(), p[i].c = read();
	int ans = solve();
	For(i, 1, n) swap(p[i].a, p[i].b);
	upd(ans, solve());
	printf("%d\n", ans);
	return 0;
}

G.篮球校赛

签到题，直接状压dp，时间复杂度 $O(5∗25n)\mathcal{O(5* 2^5 n)}$

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 100010;

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

LL dp[N][32], a[N][5];

int main() {
	int n = read();
	For(i, 1, n) For(j, 0, 4) a[i][j] = read();
	For(i, 1, n) {
		For(j, 0, 31) {
			For(k, 0, 4) {
				if ((j >> k) & 1) dp[i][j] = max(dp[i - 1][j ^ (1 << k)] + a[i][k], dp[i][j]);
				dp[i][j] = max(dp[i - 1][j], dp[i][j]);
			}
		}
	}
	printf("%lld\n", dp[n][31]);
	return 0;
}

H.分配学号

签到题，先排序确定最终的学号集合，易证明集合是唯一的，从大到小枚举可能，运用乘法原理和加法原理求得答案

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 100010;
inline void upd(int &x, int y) {x += y; x -= x >= MOD ? MOD : 0;}
inline int dec(int x, int y) {x -= y; x += x < 0 ? MOD : 0; return x;}
inline int add(int x, int y) {x += y; x -= x >= MOD ? MOD : 0; return x;}
inline int mul(int x, int y) {return 1LL * x * y % MOD;}

LL a[N], b[N];
int ans = 1, cnt[N];

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

int main() {
	int n = read();
	For(i, 1, n) a[i] = read();
	sort(a + 1, a + n + 1);
	memcpy(b, a, sizeof b);
	For(i, 2, n) {
		if (a[i] <= a[i - 1]) a[i] = a[i - 1] + 1, cnt[i] = cnt[i - 1] + 1;
	}
	int now = 0, maxi = 0, ret = 1;
	Rep(i, n, 1) {
		if (!cnt[i]) {
			ans = mul(ans, ret);
			now = maxi = 0;
			continue;
		}
		if (!now) now = cnt[i], maxi = a[i], ret = 1;
		ret = mul(ret, maxi - b[i] + 1 - now + cnt[i]);
	}
	printf("%d\n", ans);
	return 0;
}

I.Gree的心房

签到题，留出最短的一条路径，可见障碍物最多填充 $(n−1)×(m−1)(n-1)\times(m-1)$ 个位置，超过这个值就输出-1，否则输出n+m-2

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#define Mp make_pair
#define Pi pair<int, int>
#define Pb push_back
#define Fi first
#define Se second
#define Gc getchar
#define LL long long
#define Res register int
#define For(i, l, r) for(int i = (int)(l); i <= (int)(r); ++i)
#define Rep(i, r, l) for(int i = (int)(r); i >= (int)(l); --i)

using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;

inline LL read(){
	LL x = 0;
	char ch = Gc();
	bool positive = 1;
	for (; !isdigit(ch); ch = Gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = Gc()) x = (x << 1) + (x << 3) + (ch ^ 48);
	return positive ? x : -x;
}

int main() {
	int n = read(), m = read(), k = read();
	LL all = 1LL * (n - 1) * (m - 1);
	if (k <= all) printf("%d\n", n + m - 2);
	else puts("-1");
	return 0;
}