AtCoder Beginner Contest 225 (A,B,C,D,E,F)_随笔

AtCoder Beginner Contest 225 (A,B,C,D,E,F) A. Distinct Strings

三个字母的全排列得到的字符串的种类数
ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

int main() {
	
	string s;
	cin >> s;
	if(s[0] == s[1] && s[1] == s[2]) cout << 1 << endl;
	else if(s[0] == s[1] || s[0] == s[2] || s[1] == s[2]) cout << 3 << endl;
	else cout << 6 << endl;		
	
	return 0;

}

B. Star or Not

给 n-1 条边，询问是否为菊花图。
判断入度即可。
ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

int in[maxn];

int main() {
	
	int n;
	cin >> n;
	int u, v;
	for(int i = 2; i <= n; i++){
		cin >> u >> v;
		in[u]++;
		in[v]++;
	}
	
	int res = 0;
	for(int i = 1; i <= n; i++) if(in[i] == n-1) res = 1;
	
	if(res) cout << "Yes" << endl;
	else cout << "No" << endl;
		
	return 0;

}

C. Calendar Validator

判断是否为合法的矩阵。
先判断第一行是否合法，第二行开始判断是否和上一行相差 7。
ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e4 + 5;
const int N=1005;
using namespace std;

int a[maxn][10];

int main() {
	
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= m; j++){
			cin >> a[i][j];
		}
	}
	int res = 1;
	for(int i = 2; i <= m; i++){
		int x = a[1][i-1] % 7; if(x == 0) x = 7;
		int y = a[1][i] % 7; if(y == 0) y = 7;
		if(x != y-1 || a[1][i-1] != a[1][i]-1) res = 0;
	} 
	for(int i = 2; i <= n; i++){
		for(int j = 1; j <= m; j++){
			if(a[i-1][j] != a[i][j]-7) res = 0;
		}
	}
	
	if(res) cout << "Yes" << endl;
	else cout << "No" << endl;
	
	return 0;

}

D. Play Train

模拟链表即可。
ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

int pre[maxn];
int suf[maxn];

int main() {
	
	int n, m;
	cin >> n >> m;
	for(int i = 1; i <= n; i++) pre[i] = suf[i] = -1;
	int op, x, y;
	for(int i = 1; i <= m; i++){
		cin >> op >> x;
		if(op == 1){
			cin >> y;
			pre[y] = x;
			suf[x] = y;
		}
		else if(op == 2){
			cin >> y;
			pre[y] = -1;
			suf[x] = -1;
		}
		else{
			int head = x, cnt = 1;
			while(pre[head] != -1) head = pre[head], cnt++;
			int tmp = x;
			while(suf[tmp] != -1) tmp = suf[tmp], cnt++;
			cout << cnt << " ";
			while(head != -1) cout << head << " ", head = suf[head];
			cout << endl;
		}
	}
		
	return 0;

}

E. 7

几何题，偏思维。

由图可知：从x轴开始选取，当发生覆盖时，优先选取斜率较小的点比较优。
如图，1 和 2 发生覆盖，选择 1 的同时可以选择 3。
注意精度问题，用 long double。
ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

typedef struct Node{
	long double l;
	long double r;
} node;

bool cmp(node A, node B){
	return A.l < B.l;
}

node p[maxn];

int main() {
	
	int n;
	cin >> n;
	long double x, y;
	for(int i = 1; i <= n; i++){
		cin >> x >> y;
		p[i].r = (y-1) / x; // 靠下 
		if(x == 1) p[i].l = 1e9; // 靠上 
		else p[i].l = y / (x-1); 
	}
	
	sort(p+1, p+1+n, cmp); // 按照斜率排序
	int res = 0; long double l = -1e9;
	for(int i = 1; i <= n; i++){
		if(p[i].r < l) continue; // 发生覆盖不选取
		res++;
		l = max(l, p[i].l);
	}
	cout << res << endl;
	
	return 0;
}

F. String Cards

DP。
在 n 个字符串中选取 k 个任意连接，得到字典序最小的字符串。
按照 A+B > B+A 的顺序排序。
设 d p i , j dp_{i,j} dpi,j 表示前 i 个字符串取 j 个得到的最小字符。

d p i + 1 , j = d p i , j dp_{i+1,j} = dp_{i, j} dpi+1,j=dpi,j d p i + 1 , j = m i n ( d p i + 1 , j , s + d p i , j − 1 ) ， s 表示可选择的字符串 dp_{i+1,j} = min(dp_{i+1,j} , s + dp_{i, j-1})，s 表示可选择的字符串 dpi+1,j=min(dpi+1,j,s+dpi,j−1)，s表示可选择的字符串

ACcode

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod=1e9 + 7;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

string s[100], dp[55];

bool cmp(string a, string b){
	return a+b > b+a;
}

int main() {

	int n, k;
	cin >> n >> k;
	for(int i = 1; i <= n; i++) cin >> s[i];
	sort(s+1, s+n+1, cmp);
	
	for(int i = 1; i <= n; i++) dp[i] = "{";
	dp[0] = "";
	for(int i = 1; i <= n; i++){
		for(int j = k-1; j >= 0; j--){
			dp[j+1] = min(dp[j+1], s[i] + dp[j]);
		}
	}
	
	cout << dp[k] << endl;
	
	return 0;
}

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AtCoder Beginner Contest 225 (A,B,C,D,E,F)

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