东华大学2020年程序设计竞赛（同步赛）A-D F G

东华大学2020年程序设计竞赛
A

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 100+10;
const int mod = 1e9+7;

int n;
struct node{
	int index;
	ll w;
	bool operator < (const node e) const{
		if(w == e.w) return index < e.index;
		else return w > e.w;
	}
}e[N];

int main(){
	IO;
	cin >> n;
	int index;
	ll a, b, c;
	for(int i = 0; i < n; i++){
		cin >> index >> a >> b >> c;
		e[i].index = index;
		e[i].w = a + 2*b + c*3;
	}
	sort(e, e+n);
	cout << e[0].index << " " << e[0].w << "\n";
	return 0;
}

B

exgcd求$ax=1(mod)p$，当$a$是$p$的倍数时无解

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

int t;
ll a, p;

ll exgcd(ll a, ll b, ll &x, ll &y){
	if(b == 0){
		x = 1; y = 0;
		return b;
	}
	ll res = exgcd(b, a%b, x, y);
	ll tmp = x;
	x = y;
	y = tmp - a/b*y;
	return res;
}

int main(){
	IO;
	cin >> t;
	while(t--){
		cin >> a >> p;
		if(a%p == 0){
			cout << "-1\n";
		}
		else{
			ll x, y;
			exgcd(a, p, x, y);
			while(x < 0){
				x += p;
				y -= a;
			}
			cout << x%p << "\n";
		}
	}
	return 0;
}

C 最短路

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e6+10;
const int mod = 1e9+7;

int n, m;
struct Edge{
	int to, next;
}e[4*N];
int head[N], tot = 0;
void addEdge(int u, int v){
	e[tot] = (Edge){v, head[u]};
	head[u] = tot++;
}
int d[N], vis[N];
priority_queue< pair<int, int> > q;
void Dij(){
	for(int i = 0; i <= n; i++){
		d[i] = 1e9; vis[i] = 0;
	}
	d[1] = 0;
	q.push(mk(0, 1));
	while(!q.empty()){
		int u = q.top().second;
		q.pop();
		if(vis[u]) continue;
		vis[u] = 1;
		for(int i = head[u]; i != -1; i = e[i].next){
			int v = e[i].to;
			if(d[v] > d[u] + 1){
				d[v] = d[u] + 1;
				q.push(mk(-d[v], v));
			}
		}
	}
}

ll qpow(ll a, int x){
	ll res = 1;
	while(x){
		if(x&1) res = res * a % mod;
		a = a * a % mod;
		x >>= 1;
	}
	return res;
}

int main(){
	IO;
	cin >> n >> m;
	memset(head, -1, sizeof(int)*(n+1));
	int u, v;
	for(int i = 0; i < m; i++){
		cin >> u >> v;
		addEdge(u, v);
		addEdge(v, u);
	}
	Dij();
	ll ans = 0;
	for(int i = 2; i <= n; i++){
		ans = (ans + qpow(2, d[i])) % mod;
	}
	cout << ans << "\n";
	return 0;
}

D

暴力模拟，每次选最小堆加

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 100+10;
const int mod = 1e9+7;

int t, n, a[N];

int main(){
	IO;
	cin >> t;
	while(t--){
		cin >> n;
		int sum = 0;
		for(int i = 0; i < n; i++){
			cin >> a[i];
			sum += a[i];
		}
		if(sum%n != 0){
			cout << "No\n";
			continue;
		}
		sort(a, a+n);
		int flag = 1;
		for(int i = 1; i < n; i++){
			if((a[i]-a[0]) % n != 0){
				flag = 0;
			}
		}
		int x = sum/n;
		while(flag){
			int cnt1 = 0, cnt2 = 0;
			for(int i = 0; i < n; i++){
				if(a[i] == x) cnt1++;
				if(a[i] == 0) cnt2++;
			}
			if(cnt1 == n) {
				flag = 1; break;
			}
			if(cnt2 > 1){
				flag = 0; break;
			}
			sort(a, a+n);
			a[0] += n-1;
			for(int i = 1; i < n; i++){
				a[i]--;
			}
		}
		if(flag == 0) {
			cout << "No\n";
		}
		else cout << "Yes\n";
	}
	return 0;
}

F

连续的0变为1，然后1又可以变为0，所以连续的0贡献2步，对结果没有影响。对每个字符串我们只需要统计1的个数就好。
当所有串含1的个数都是偶数时，无论你选择对哪些串进行操作，对方只要也选这些串进行操作，就能继续保证所有串含1的个数都是偶数。由于是Alice先动，所以只有当所有串含1的个数都是偶数时，Alice才会输，其他情况则是Bob落败。

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

int t, n;
string s;

int main(){
	IO;
	cin >> t;
	while(t--){
		cin >> n;
		int ans = 0;
		for(int i = 0; i < n; i++){
			cin >> s;
			int num = 0;
			for(int j = 0; j < s.size(); j++){
				if(s[j] =='1') num++;
			}
			if(num % 2) ans++;
		}
		if(ans) cout << "sdzNB\n";
		else cout << "kgNB\n";
	}
	return 0;
}

G

最多只有五个数连乘，$-1\ast 1\ast 1\ast 1\ast -1$，然后dp

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

int t, n, a[N], dp[N];

int main(){
	IO;
	cin >> t;
	while(t--){
		cin >> n;
		for(int i = 0; i <= n; i++){
			dp[i] = -1e9;
		}
		for(int i = 1; i <= n; i++){
			cin >> a[i];
		}
		dp[0] = 0;
		for(int i = 1; i <= n; i++){
			int x = 1;
			for(int  j = 1; j <= 5; j++){
				if(i - j < 0) continue;
				x *= a[i-j+1];
				dp[i] = max(dp[i], dp[i-j] + x);
			}
		}
		cout << dp[n] << "\n";
	}
	return 0;
}