第十六届北京师范大学程序设计竞赛决赛（网络同步赛）

题目链接  第十六届北京师范大学程序设计竞赛决赛

一句话总结：迟到选手抢到FB之后进入梦游模式最后因为忘加反向边绝杀失败……

好吧其实还是自己太弱

下面进入正题

Problem A

签到题（读题是一件非常有趣事情）

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;


int main(){

	int T; string s;
	scanf("%d", &T);

	while (T--){
		int n;
		scanf("%d", &n);
		int fg = 1;
		rep(i, 1, n){
			cin >> s;
			if (s != "PERFECT") fg = 0;
		}
		puts(fg ? "MILLION Master" : "NAIVE Noob");
	}


	return 0;
}

 

Problem B

设读进来的那个序列为$b_{i}$

要还原出的那个序列答案为$a_{i}$

我们求出$a_{i}$对$b_{i}$每一项的贡献就可以了。

这个过程实现起来不难。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)

typedef long long LL;

const LL mod = 1e9 + 7;
const int N  = 1e3 + 10;

int T;
int n;
LL  a[N], b[N], c[N];
LL  k;

inline LL Pow(LL a, LL b, LL mod){
        LL ret(1);
        for (; b; b >>= 1, (a *= a) %= mod) if (b & 1) (ret *= a) %= mod;
        return ret;
}

void pre(){
	int cnt = 0;
	c[++cnt] = 1;

	LL now = 1;
	rep(i, 1, n - 1){
		now = now * (k + i - 1) % mod;
		now = now * Pow((LL)i, mod - 2, mod) % mod;
		c[++cnt] = now;
	}
}

int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%lld", &n, &k);
		memset(c, 0, sizeof c);
		pre();

		rep(i, 1, n) scanf("%lld", b + i);
		a[1] = b[1];
		rep(i, 1, n){
			int cnt = 0;
			rep(j, i, n){
				++cnt;
				b[j] -= (c[cnt] * a[i] % mod);
				b[j] += mod;
				b[j] %= mod;
			}

			a[i + 1] = b[i + 1];
		}

		rep(i, 1, n - 1) printf("%lld ", a[i]);
		printf("%lld
", a[n]);
	}


	return 0;
}

　　

 

Problem C

打表之后找规律

#include<bits/stdc++.h>

using namespace std;

int T;
int n;

int main(){

	cin >> T;
	while (T--){
		cin >> n;
		cout << fixed << setprecision(10) << (n * n - 1) / 3.0 << endl;
	}
	return 0;
}

 

Problem D

这个DP转移

每个点有两种转移的方向，每个方向取最近的那个点就好了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 1e5 + 10;

int T;
int n, m, k;
LL a[N], b[N], c[N], f[N], g[N], h[N];
map <LL, LL> mp;
set <LL> s;

int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%d%d", &n, &m, &k);
		rep(i, 1, n) scanf("%lld", a + i);
		rep(i, 1, m) scanf("%lld", b + i);
		rep(i, 1, k) scanf("%lld", c + i);

		rep(i, 1, n) f[i] = 1;
		rep(i, 1, m) g[i] = 1e15;

		mp.clear();
		rep(i, 1, n) mp[a[i]] = f[i];
		mp[1e18] = 1e18, mp[-1e18] = 1e18;

		s.clear();

		rep(i, 1, n) s.insert(a[i]);
		s.insert(1e18);
		s.insert(-1e18);

		rep(i, 1, m){
			LL xx = b[i];
			auto it = s.lower_bound(xx);
			g[i] = min(g[i], mp[*it] + abs((*it) - xx) + 1);
			--it;
			g[i] = min(g[i], mp[*it] + abs((*it) - xx) + 1);
		}

		rep(i, 1, k) h[i] = 1e15;
		mp.clear();
		rep(i, 1, m) mp[b[i]] = g[i];
		mp[1e18] = 1e18, mp[-1e18] = 1e18;


		s.clear();
		rep(i, 1, m) s.insert(b[i]);
		s.insert(1e18);
		s.insert(-1e18);

		rep(i, 1, k){
			LL xx = c[i];
			auto it = s.lower_bound(xx);
			h[i] = min(h[i], mp[*it] + abs((*it) - xx) + 1);
			--it;
			h[i] = min(h[i], mp[*it] + abs((*it) - xx) + 1);
		}

		LL ans = 1e18;
		rep(i, 1, k) ans = min(ans, h[i]);
		printf("%lld
", ans);
	}

	return 0;
}

 

 

Problem E

考虑到长度为$k$的子序列个数不超过$10^{5}$，那么直接暴力，搜索深度不超过$9$。

注意：$k$比较大的时候考虑反面即可。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 1e5 + 10;

int T;
int n, k;
int z;
int c[N];
LL  a[N];
LL  ans;

void check(LL x){
	LL vv = x * x;
	ans ^= vv;
}

void dfs(int x, LL now){
	if (x > k){
		check(now);
		return;
	}

	else{
		rep(i, z, n){
			int la = z;
			z = i + 1;
			dfs(x + 1, now + a[i]);
			z = la;
		}
	}
}

void dfs2(int x, LL now){
	if (x > k){
		check(now);
		return;
	}

	else{
		rep(i, z, n){
			int la = z;
			z = i + 1;
			dfs2(x + 1, now - a[i]);
			z = la;
		}
	}
}


int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%d", &n, &k);
		LL sum = 0;
		rep(i, 1, n) scanf("%lld", a + i), sum += a[i];
		z = 1;
		ans = 0;
		if (k > n - k){
			k = n - k;
			dfs2(1, sum);
		}
		else{
			dfs(1, 0);
		}
		printf("%lld
", ans);
	}


	return 0;
}

 

 

Problem F

显然答案是单调的。

二分答案，把那些符合条件的汤圆一个个加入队列，然后BFS，到最后如果所有汤圆都被删除了，那么该答案可行。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;
typedef pair <int, LL> PII;

const int N = 1e5 + 10;

int T;
int n, m;
int vis[N], inq[N];
LL a[N], bb[N];
LL l, r;
vector <PII> v[N];


bool check(LL now){
	queue <int> q;
	memset(vis, 0, sizeof vis);
	memset(inq, 0, sizeof inq);
	rep(i, 1, n) bb[i] = a[i];
	rep(i, 1, n) if (bb[i] <= now) q.push(i), inq[i] = 1;
	
	while (!q.empty()){
		int x = q.front();
		vis[x] = 1;
		q.pop();
		inq[x] = 0;

		for (auto edge : v[x]){
			int u = edge.fi;
			LL  w = edge.se;
			if (vis[u]) continue;
			bb[u] -= w;
			if (bb[u] <= now){
				if (!inq[u]) q.push(u), inq[u] = 1;
			}
		}
	}

	rep(i, 1, n) if (!vis[i]) return false;
	return true;
}

int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%d", &n, &m);
		rep(i, 0, n + 1) v[i].clear();
		rep(i, 0, n + 1) a[i] = 0;
		rep(i, 1, m){
			int x, y;
			LL z;
			scanf("%d%d%lld", &x, &y, &z);
			v[x].push_back({y, (LL)z});
			v[y].push_back({x, (LL)z});
			a[x] += z;
			a[y] += z;
		}

		l = 0, r = 1e14;

		while (l + 1 < r){
			LL mid = (l + r) / 2ll;
			if (check(mid)) r = mid;
			else l = mid + 1;
		}

		if (check(l)) printf("%lld
", l);
		else printf("%lld
", r);
	}

	return 0;
}

 

Problem G

模拟题，没什么好说的。（我还是WA了好几发）

#include <bits/stdc++.h>

using namespace std;

#define	rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define	dec(i, a, b)	for (int i(a); i >= (b); --i)
#define	MP		make_pair
#define	fi		first
#define	se		second

typedef long long LL;

char s[24];
bool ff[24];
int T;

bool upper(char c){ return c >= 'A' && c <= 'Z'; }

bool check(int n){
	bool re = 1;
	bool flag = 0;
	int cnt = 0;
	if (!upper(s[0])) s[0] += 'A' - 'a', flag = 1;

	rep(i, 1, n - 1){
		if (upper(s[i]) && upper(s[i - 1])) re = 0;
		if (upper(s[i])) cnt++, ff[i] = 1;
	}

	if (flag) s[0] += 'a' - 'A';
	return re && (cnt) && !upper(s[n - 1]);
}


int main(){


	cin >> T;

	while (T--){
		scanf("%s", &s);
		memset(ff, 0, sizeof ff);
		int n = strlen(s);
		if (check(n)){
			rep(i, 0, n - 1){
				if (ff[i]) putchar('_');
				if (upper(s[i])) s[i] += 'a' - 'A';
				putchar(s[i]);
			}
			putchar(10);
		}
		else puts(s);
	}

	return 0;
}

Problem H

数据结构题，留坑。

Problem I

每次交换相邻的两个不一样的元素会使整个序列的逆序对数改变$1$

那么计算一下逆序对个数就好了。

#include <bits/stdc++.h>

using namespace std;

#define rep(i, a, b)	for (int i(a); i <= (b); ++i)
#define dec(i, a, b)	for (int i(a); i >= (b); --i)
#define MP		make_pair
#define fi		first
#define se		second


typedef long long LL;

const int N = 1e6 + 10;

int T;
int n;
int a[N];
char s[N];
LL  k;


int main(){

	scanf("%d", &T);
	while (T--){
		scanf("%d%lld", &n, &k);
		scanf("%s", s + 1);

		rep(i, 1, n) a[i] = (s[i] == 'D');
		LL cc = 0;
		rep(i, 1, n) cc += a[i];
		LL ju = 1ll * cc * (LL)(n - cc);
		if (ju < k) {
			puts("-1");
			continue;
		}

		LL cnt = 0, ss = 0;
		rep(i, 1, n){
			if (a[i] == 0) cnt += ss;
			ss += a[i];
		}

		printf("%lld
", abs(cnt - k));
	}


	return 0;
}

Problem J

计算几何，留坑。

Problem K

留坑。