牛客小白月赛83(A-E)

牛客小白月赛83(A-E)

A 小天的金银铜铁

代码

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// Problem: 小天的金银铜铁
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/72041/A
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define sc second
#define endl '\n'
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define debug(x) cout<<"----Line#"<<x<<"----"<<endl
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
typedef pair<char, char> PCC;
typedef pair<int, string> PIS;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const ll mod = 1e9 + 7;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 1010;

void solve () {
    int a, b, c, d, e;
    cin >> a >> b >> c >> d >> e;
    int n, m, x, y;
    cin >> n >> m >> x >> y;
	if (n * a + m * b + c * x + d * y >= e) {
		cout << "YES" << endl;
	} else {
		cout << "NO" << endl;
	}
    return ;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    //cin >> T;
    while (T --) {
        solve ();
    }
    return 0;
}

B 小天的魔法

分析

sort排序之后贪心即可

点击查看代码

// Problem: 小天的魔法
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/72041/B
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define sc second
#define endl '\n'
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define debug(x) cout<<"----Line#"<<x<<"----"<<endl
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
typedef pair<char, char> PCC;
typedef pair<int, string> PIS;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const ll mod = 1e9 + 7;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 1010;
int n, m, x;
int a[N], b[N];

void solve () {
    cin >> n >> m >> x;
    for (int i = 1; i <= n; i ++) cin >> a[i];
    for (int i = 1; i <= n; i ++) cin >> b[i];
    sort(a + 1, a + n + 1);
    sort(b + 1, b + n + 1);
    ll s = 0;
    int cn = 0;
    for (int i = n; i >= 1; i --) {
    	if (s + b[i] >= x) {
    		cn ++;
    		cout << cn << endl;
    		return ;
    	}
    	s += a[i] * b[i];
    	if (s >= x) {
    		cn += 2;
    		cout << cn << endl;
    		return ;
    	}
    	cn += 2;
    }
    cout << -1 << endl;
    return ;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    //cin >> T;
    while (T --) {
        solve ();
    }
    return 0;
}

C 小天的 Minecraft

分析

一共只有三种达成目标的情况，分别是16铜，12铜4银，12铜4金

点击查看代码

// Problem: 小天的 Minecraft
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/72041/C
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define sc second
#define endl '\n'
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define debug(x) cout<<"----Line#"<<x<<"----"<<endl
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
typedef pair<char, char> PCC;
typedef pair<int, string> PIS;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const ll mod = 1e9 + 7;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 1010;

void solve () {
    int a, b, c; cin >> a >> b >> c;
    double ans = 1820 * pow(a * 1.0 / 16, 12) * (pow(b * 1.0 / 16, 4) + pow(c * 1.0 / 16, 4)) + pow(a * 1.0 / 16, 16);
    printf("%.10lf\n", ans);
    return ;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T --) {
        solve ();
    }
    return 0;
}

D 小天的子序列

分析

如果使用暴力求解，我们可以通过枚举区间加组合数的方式n^2求出题目要求的子序列数量。但是会TLE
使用mp数组，mp[x][y][l]意义为以x开头，y结尾，长度为l的子串的数量，可以通过初始化mp数组来求出答案。在每个询问时，我们已经知道了以x开头y结尾，长度为len~n的所有子串的数量，因此可以O(n)求出答案

点击查看代码

// Problem: 小天的子序列
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/72041/D
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define sc second
#define endl '\n'
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define debug(x) cout<<"----Line#"<<x<<"----"<<endl
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
typedef pair<char, char> PCC;
typedef pair<pair<char, char>, int> PCCI;
typedef pair<int, string> PIS;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const ll mod = 998244353;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 510;
int n, m, c[N][N];
int mp[26][26][N];

void init() {
    for(int i = 0; i < N; i ++) {
        for(int j = 0; j <= i; j ++) {
            if (!j)
                c[i][j] = 1;
            else {
                c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
            }
        }
    }
}

void solve () {
	cin >> n;
	string s; cin >> s;
	for (int i = 0; i < n; i ++) {
		for (int j = i + 1; j < n; j ++) {
			mp[s[i] - 'a'][s[j] - 'a'][j - i + 1] ++;
		}
	}
	cin >> m;
	while (m --) {
		char x, y; cin >> x >> y;
		int len; cin >> len;
		ll ans = 0;
		for (int i = len; i <= n; i ++) {
			ans = (ans + mp[x - 'a'][y - 'a'][i] * c[i - 2][len - 2]) % mod;
		}
		cout << ans << endl;
	}
    return ;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    init();
    //cin >> T;
    while (T --) {
        solve ();
    }
    return 0;
}

E 小天的贪吃蛇

分析

可以将格子转化为序列
我们考虑使用set维护每个字符的所有下标，因此对于每一个2/3类型的询问{x,y}，首先将其转化为一维下标pos，然后可以对除了此字符以外的所有字符的set进行二分，求出pos之后第一个出现的位置的最小值，就是最终的答案

点击查看代码

// Problem: 小天的贪吃蛇
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/72041/E
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>

using namespace std;

#define fi first
#define sc second
#define endl '\n'
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define meminf(a) memset(a,0x3f,sizeof(a))
#define debug(x) cout<<"----Line#"<<x<<"----"<<endl
typedef long long ll;
typedef double db;
typedef pair<int, int> PII;
typedef pair<char, char> PCC;
typedef pair<int, string> PIS;
const int INF = 0x3f3f3f3f;
const double eps = 1e-6;
const ll mod = 1e9 + 7;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}

const int N = 1e5 + 10;
int n, m, q;
string col, row;
map<PII, int> mp1, mp2;
set<int> s1[26], s2[26];

void solve () {
    cin >> n >> m;
    vector s(n + 1, vector<char> (m + 1));
    for (int i = 1; i <= n; i ++) {
    	for (int j = 1; j <= m; j ++) {
    		cin >> s[i][j];
    	}
    }
    for (int i = 1, v = 1; i <= n; i ++)
    	if (i & 1) 
    		for (int j = 1; j <= m; j ++) 
    			mp1[{i, j}] = v ++;
    	else 
    		for (int j = m; j >= 1; j --)
    			mp1[{i, j}] = v ++;
    			
    for (int j = 1, v = 1; j <= m; j ++) 
    	if (j & 1) 
    		for (int i = 1; i <= n; i ++) 
    			mp2[{i, j}] = v ++;
     	else 
    		for (int i = n; i >= 1; i --) 
    			mp2[{i, j}] = v ++;
	
    for (int i = 1; i <= n; i ++) 
    	for (int j = 1; j <= m; j ++) {
    		int ch = s[i][j] - 'a';
    		s1[ch].insert(mp1[{i, j}]);
    		s2[ch].insert(mp2[{i, j}]);
    	}
 	cin >> q;
    while (q --) {
    	int op, x, y; cin >> op >> x >> y;
    	if (op == 1) {
    		char c; cin >> c;
    		char ch = s[x][y];
    		s1[ch - 'a'].erase(mp1[{x, y}]);
    		s2[ch - 'a'].erase(mp2[{x, y}]);
    		s1[c - 'a'].insert(mp1[{x, y}]);
    		s2[c - 'a'].insert(mp2[{x, y}]);
    		s[x][y] = c;
    	} else if (op == 2) {
    		char ch = s[x][y];
    		int ans = INF;
    		for (char c = 'a'; c <= 'z'; c ++) {
    			if (c == ch) continue;
    			else {
    				auto pos = s1[c - 'a'].lower_bound(mp1[{x, y}]);
    				if (pos == s1[c - 'a'].end()) {
    					ans = min(ans, n * m - mp1[{x, y}] + 1);
    				} else {
    					ans = min(ans, *pos - mp1[{x, y}]);
    				}
    			}
    		}
    		cout << ans << endl;
    	} else {
    		char ch = s[x][y];
    		int ans = INF;
    		for (char c = 'a'; c <= 'z'; c ++) {
    			if (c == ch) continue;
    			else {
    				auto pos = s2[c - 'a'].lower_bound(mp2[{x, y}]);
    				if (pos == s2[c - 'a'].end()) {
    					ans = min(ans, n * m - mp2[{x, y}] + 1);
    				} else {
    					ans = min(ans, *pos - mp2[{x, y}]);
    				}
    			}
    		}
    		cout << ans << endl;
    	}
    }
    return ;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int T = 1;
    //cin >> T;
    while (T --) {
        solve ();
    }
    return 0;
}

F 小天的 A+B

明天再补