牛客小白月赛56

A 阿宁的柠檬

signed main() {
    int a = read(), b = read(), n = read();
    LL mi = 1 * n;
    LL ma = a * n + b * n;
    cout << mi << ' ' << ma << endl;
	return 0;
}

B 阿宁与猫咪

int main() {
    int m = read();
    if(m < 2) cout << 1 << endl;
    else {
        cout << 2 << endl;
    }
    for(int i = 0; i < m; i++) cout << 1 << ' ';
    cout << endl;
	return 0;
}

C 阿宁吃粽子

思路：

从小到大按1-9、11-19...这样排。

代码：

int main() {
	int n = read();
	for(int i = 1; i <= n; i++) w[i] = read();
	sort(w + 1, w + 1 + n);
	int cnt = n / 10;
	int k = 1;
	for(int i = 1; i <= cnt; i++) ans[i*10] = w[k++];
	if(n % 10 == 0) {
		for(int i = 1; i <= 9; i++) {
			for(int j = 0; j < cnt; j++) {
				ans[i + j * 10] = w[k++];
			}
		}
	}else {
		for(int i = 1; i <= 9; i++) {
			for(int j = 0; j <= cnt; j++) {
				if(i + j * 10 > n) continue;
				ans[i + j * 10] = w[k++];
			}
		}
	}
	for(int i = 1; i <= n; i++) cout << ans[i] << ' ';
	cout << endl;
	return 0;
}

D 阿宁的质数

思路:

先预处理，把估计范围的质数全部筛出来，然后再预处理1-i的最小未出现质数，这里我当时用的是二分，但是过不了后面的数据。直接暴力++就行了。

代码：

/*
	qwq!
*/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <map>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <cmath>
#include <unordered_map>
using namespace std;
#define pb push_back
#define pu push
#define fi first
#define se second
#define LL long long
typedef pair<int,int> PII;
const int INF = 0x3f3f3f3f;
const int N = 5e6 + 10;
int w[N];
int primes[N], cnt;
int ans[N];
bool st[N];
map<int,bool>stx;

void get_primes() {
    for(int i = 2; i <= N; i++) {
        if(!st[i]) primes[cnt++] = i;
        for(int j = 0; primes[j] <= N / i; j++) {
            st[primes[j] * i] = true;
            if(i % primes[j] == 0) break;
        }
    }
}
int read () {
	int k=0,f=1;
	char c=getchar ();
	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar ();}
	while (c>='0'&&c<='9') {k=k*10+c-'0';c=getchar ();}
	return k*f;
}

bool check(int x) {
    if(x < 2) return false;
    for(int i = 2; i <= x / i; i++)
            if(x % i == 0) return false;
    return true;
}

int main() {
    int n = read(), q = read();
    for(int i = 1; i <= n; i++) w[i] = read();
    get_primes();
    int now = 0;
    ans[0] = primes[now];
    for(int i = 1; i <= n; i++) {
        if(w[i] == primes[now]) {
            now++;
            while(stx[primes[now]]) now++;
            ans[i] = primes[now];
        }else ans[i] = primes[now];
        stx[w[i]] = 1;
    }
    while(q--) {
        int x = read();
        cout << ans[x] << endl;
    }
	return 0;
}

E 阿宁睡大觉

思路：尽量让两个Z靠拢。

代码：

int main() {
    int n = read(), k = read();
    string s; cin >> s;
    s = 'x' + s;
    int cnt = 0, last = -1;
    for(int i = 1; i <= n; i++) {
    	if(s[i] == 'Z' && last == -1) {
    		last = i;
		}else if(s[i] == 'Z'){
			w[cnt++] = i - last - 1;
			last = i;
		}
	}
	sort(w, w + cnt); 
	int ans = 0;
 // for(int i = 0; i < cnt; i++) cout << w[i] << ' ';
 // cout << endl;
	for(int i = 0; i < cnt; i++) {
		if(k < w[i]) break;
		ans += 4;
		k -= w[i];
	}
	cout << ans << endl;
	return 0;
}

F 阿宁去游玩