2019 ICPC Asia Danang Regional Contest

2019 ICPC Asia Danang Regional Contest

A. Abstract Painting

妙啊~真是妙蛙种子吃着妙脆角妙进了米奇妙妙屋妙到家了

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

const int mod = 1e9+7;

int qpow(int a,int p){
    int ans = 1;
    while(p){
        if(p & 1)ans = (ans * a) % mod;
        a = a * a % mod;
        p >>= 1;
    }
    return ans;
}

signed main(){
    int T,a,b;
    cin >> T;
    while(T--){
        cin >> a >> b;
        cout << qpow(2,a*b)*qpow(3,a+b)%mod << endl;
    }
}

B. Banana Problem

题目很长，读懂题意就可以了。直接模拟。

根据特殊点倒过来找猴子。

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

int n, h, r, s;

const int maxn = 3e5 + 10;
int up[maxn], hor[maxn];

struct rope {
	int h, len, to;
	bool operator < (const rope& rhs)const {
		return h < rhs.h;
	}
};
vector<rope>P[maxn];

int main() {
	scanf("%d%d%d%d", &n, &h, &r, &s);
	for (int i = 0; i < r; i++) {
		int len, a, b, H;
		scanf("%d%d%d%d", &len, &a, &b, &H);
		P[a].push_back(rope{ H,len,b });
		P[b].push_back(rope{ H,len,a });
	}
	for (int i = 1; i <= n; i++) {
		scanf("%d%d", up + i, hor + i);
		sort(P[i].begin(), P[i].end());
	}
	int ans = 0;
	for (int i = 1; i <= s; i++) {
		int pos, nowh, x, y;
		scanf("%d%d%d%d", &pos, &nowh, &x, &y);
		
		long long up_dis = 0;
		long long hor_dis = 0;
		while (1) {
			if (P[pos].empty()) {
				up_dis += nowh;
				break;
			}
			auto p = lower_bound(P[pos].begin(), P[pos].end(), rope{ nowh,0,0 });
			if (p == P[pos].begin()) {
				up_dis += nowh;
				break;
			}
	
			rope x = *prev(p);
			up_dis += nowh - x.h;
			hor_dis += x.len;
			pos = x.to;
			nowh = x.h;
		}
		long long sum = up_dis * up[pos] + hor_dis * hor[pos];
		if (sum % (x + y) >= 1 && sum % (x + y) <= x)ans++;
	}
	printf("%d
", ans);
}

D. Dating time

将钟盘划分成 (43200) 等份，则一秒钟时针走 (1) 格，分针走 (12) 格

算出相对的初始距离，利用相对速度算出最后的相对距离。

遍历这些距离就可以。

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

int main() {
	int h1, m1, h2, m2, p;
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%d:%d %d:%d %d", &h1, &m1, &h2, &m2, &p);
		int s1 = h1 * 3600 + m1 * 60;
		int s2 = h2 * 3600 + m2 * 60;
		
		int dx = 12 * (s1 % 3600) - (s1 % 43200);
		while (dx < 43200)dx += 43200;

		int ans = 0;
		int ed = dx + (s2 - s1) * 11;
		for (int x = dx; x <= ed; x++) {
			if (p == 0) {
				if (x % 43200 == 0)ans++;
			}
			if (p == 90) {
				if (x % 43200 == 10800 || x % 43200 == 32400)ans++;
			}
			if (p == 180) {
				if (x % 43200 == 21600)ans++;
			}
		}
		cout << ans << endl;
		
	}
}

G. Generating Numbers

正常模拟就可以。

注意中间有 (1) 的情况，如果第一位大于 (1) 的话，可以省一步

(eg:) (213)

(102 ightarrow 210)

而不是

(102 ightarrow 201)

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

int main() {
	int t;
	string s;

	cin >> t;
	int cnt[20];
	cnt[1] = 0;
	for (int i = 2; i <= 10; i++) {
		cnt[i] = cnt[i - 1] + (i - 2) * 10 + 9;
	}
	while (t--) {
		cin >> s;
		if (s.size() == 1) {
			cout << s[0] - '1' << endl;
			continue;
		}
		int F = 0;
		for (int i = 1; i < s.size(); i++) {
			if (s[i] != '0') {
				F = 1;
				break;
			}
		}
		int ans = 0;
		if (!F) { //x0000
			ans = 1;
			if (s[0] == '1') {
				s = string(s.size() - 1, '9');
			}
			else {
				s = string(1, s[0] - 1) + string(s.size() - 1, '9');
			}
		}

		ans += cnt[s.size()];

		if (s.size() == 1) {
			ans += s[0] - '1';
		}
		else {
			int f = 1;
			int ff = 0;
			if (s[0] > '1') {
				ans += s[0] - '0' + 1;
				f = 0;
			}
			int one = 0;
			for (int i = 1; i < s.size() - 1; i++) {
				if (s[i] != '0') {
					ff++;
					if (!f) {
						f = 1;
						ans--;
					}
					ans += s[i] - '0';
					ans++;
				}
				if (s[i] == '1')one++;
			}
			ans += s[s.size() - 1] - '0';
			if (!f && s[s.size() - 1] > '0') {
				ans--;
			}
			if (s[0] > '1' && one)ans--;
			
		}

		cout << ans << endl;;
	}
}

H. Hanjie

二进制预处理出每行可能的数值，直接暴力即可

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

int r, c;// [1,6]
vector<int>R[10], C[10];
int Cnums[10];
vector<int>Rnums[10];
vector<int>seq;

void cal_seq(int num) {
	seq.clear();
	int f = 0;
	while (num) {
		if (num & 1) {
			if (!f) {
				seq.push_back(1);
			}
			else {
				seq[seq.size() - 1]++;
			}
			f = 1;
		}
		else {
			f = 0;
		}
		num >>= 1;
	}
	reverse(seq.begin(), seq.end());
}
void generate(int row) {
	for (int num = 0; num < (1 << c); num++) {
		cal_seq(num);
		if (seq == R[row]) {
			Rnums[row].push_back(num);
		}
	}
}

int ans = 0;
vector<int>cal;

bool judge() {
	memset(Cnums, 0, sizeof Cnums);
	for (int i = 0; i < r; i++) {
		int num = cal[i];
		int cnt = 0;
		while (num) {
			if (num & 1) {
				Cnums[c - cnt] += (1 << (r - i - 1));
			}
			cnt++;
			num >>= 1;
		}
	}
	
	for (int i = 1; i <= c; i++) {
		cal_seq(Cnums[i]);
		if (seq != C[i]) {
			return false;
		}
	}
	return true;
}
void dfs(int row) {
	if (row > r) {
		if (judge()) {
			ans++;
		}
		return;
	}
	for (int i : Rnums[row]) {
		cal.push_back(i);
		dfs(row + 1);
		cal.pop_back();
	}
}

int main() {
	cin >> r >> c;
	for (int i = 1; i <= r; i++) {
		int k; cin >> k;
		for (int j = 1; j <= k; j++) {
			int x; cin >> x;
			R[i].push_back(x);
		}
	}
	for (int i = 1; i <= c; i++) {
		int k; cin >> k;
		for (int j = 1; j <= k; j++) {
			int x; cin >> x;
			C[i].push_back(x);
		}
	}

	for (int i = 1; i <= r; i++) {
		generate(i);
	}

	dfs(1);
	
	cout << ans << endl;
}

I. Inspecting Illumination

一个交互题，二进制构造就可以了。

挺巧妙的。

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

const int maxn = 1e3 + 10;
int ans[maxn];
int q[maxn];
int re[maxn];
int cnt;
int n;

int main() {
	cin >> n;
	int m = 1;
	while ((1 << m) - 1 < n)m++;
	//cout << m << endl;

	for (int i = 1; i <= m; i++) {
		cnt = 0;
		for (int j = 1; j <= n; j++) {
			if (j & (1 << (i - 1))) {
				q[++cnt] = j;
			}
		}
		cout << "ASK " << cnt;
		for (int k = 1; k <= cnt; k++) {
			cout << " " << q[k];
		}
		cout << endl;
		cout.flush();

		for (int k = 1; k <= cnt; k++) {
			cin >> re[k];
			ans[re[k]] += (1 << (i - 1));
		}	
		
	}

	
	cout << "ANSWER";
	for (int i = 1; i <= n; i++) {
		cout << " " << ans[i];
	}
	cout.flush();
}

L. Latin Square

二分图匹配

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

int A[200][200];
int vis[200];
int nvis[200];
int match[200];
int n, k;
bool dfs(int x) {

	for (int i = 1; i <= n; i++) {
		if (vis[i] || A[x][i] != 0)continue;
		vis[i] = 1;
		if (!match[i] || dfs(match[i])) {
			match[i] = x;
			return true;
		}
	}
	return false;
}
int main() {
	cin >> n >> k;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cin >> A[i][j];
			nvis[A[i][j]] = 1;
		}
	}
	for (int i = 1; i <= n; i++) {
		if (nvis[i])continue;

		memset(match, 0, sizeof match);
		for (int j = 1; j <= n; j++) {
			memset(vis, 0, sizeof vis);
			dfs(j);
		}

		for (int j = 1; j <= n; j++) {
			A[match[j]][j] = i;
		}
	}
	cout << "YES" << endl;
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= n; j++) {
			cout << A[i][j] << " 
"[j == n];
		}
	}
}

M. Moscow Dream

n >= 3

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

int main(){
    int a,b,c,n;
    cin >> a >> b >> c >> n;
    string ans[2] = {"NO","YES"};
    cout << ans[a>0&&b>0&&c>0&&a+b+c>=n&&n>=3] << endl;
    
}