2.2

2.2

[P2527 SHOI2001] Panda的烦恼 - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 发现n小k大，复杂度为O（n * k），大概会吃到1e7的复杂度，姑且可以低空飘过

• 考虑运用set，java中的话就是TreeSet，自动提供排序和排重的功能

• ok，那么先放入一个1，然后根据p依据放入set

• 那么以下有几个问题‘

• • 1 WA，原因是没开long，okSet<Long>
  • 2 MLE, 原因是如果set开long，超出空间要求了，改进为只要超过2147483647就跳过，int范围$2^{31} - 1 $ ,那么问题解决
  • 3 TLE，考虑到可能加入了许多已经超过k号的数，所以进行优化

• 建议本题用cpp跑，java太慢了，容易被卡常数

#include<bits/stdc++.h>
using namespace std;
int n,k,f=1,a[105];long long mx,INF=2000000000;set<long long>s;
//f表示当前set中是否有k个数
int main()
{
	scanf("%d%d",&n,&k),s.insert(1);
	for(int i=1;i<=n;i++) scanf("%d",a+i),a[i]==1?(i--,n--):0;
	sort(a+1,a+n+1),mx=1;//最大值变成当前唯一的一个1
	for(int i=1;i<=k;i++)
	{
		long long t=*s.begin();s.erase(s.begin());
		for(int i=1;i<=n&&t*a[i]<=INF;i++) if(f||a[i]*t<=mx) s.insert(t*a[i]),mx=max(mx,t*a[i]);
		//首先，刚刚加的优化仍然有（在循环条件中）
		//然后，这个判断条件就是刚刚说的那个break的反条件QAQ
		f=((int)s.size()+i<=k);
		//更新f，检查加入后是否set中有k个数
	}
	return printf("%lld\n",*s.begin()),0;
}

• 贴个90分java代码

package shuati;

import java.io.*;
import java.util.*;

public class Main implements Runnable {

    static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static int n, k;
    static int[] p = new int[105];
    static int ans;
    static TreeSet<Integer>set = new TreeSet<>();
    static int INX = 2147483647;

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 1 << 29).start();
    }

    public static int nextInt() throws IOException {
        cin.nextToken();
        return (int) cin.nval;
    }

    @Override
    public void run() {
        try {
            n = nextInt(); k = nextInt();
            for(int i = 1; i <= n; i++) p[i] = nextInt();

            set.add(1);
            int mx = 1;
            boolean f = true;
            for(int i = 1; i <= k; i++) {
                int begin = set.first(); set.remove(set.first());
                for(int j = 1; j <= n; j++) {
                    if((long)begin * p[j] > INX) continue;
                    if(f || p[j] * begin <= mx) {
                        set.add(p[j] * begin);
                        mx=Math.max(mx,begin * p[j]);
                    }
                    set.add(begin * p[j]);
                }
                f=((int)set.size() + i <= k);
            }
            cout.println(set.first());
            cout.flush();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
    }

}

Smallest number - 洛谷 | 计算机科学教育新生态 (luogu.com.cn)

• 爆搜就完了，还有就是可以全排列，因为运算符要按顺序，只用全排数字，一共就4个数字，只有24种可能性

import java.io.*;
import java.util.*;

public class Main implements Runnable {

//    static StreamTokenizer cin = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static Scanner cin = new Scanner(System.in);
//    static BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter cout = new PrintWriter(new OutputStreamWriter(System.out));
    static long[] arr = new long[5];
    static char[] t = new char[4];
    static boolean[] str = new boolean[5];
    static long ans = 1000000000000000L;

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 1 << 29).start();
    }

    public static void dfs(int cnt) {
        if (cnt == 4) {
            for (int i = 1; i <= 4; i++) {
                if (!str[i]) {
                    long s = arr[i];
                    ans = Math.min(s, ans);
                    return;
                }
            }
        }

        for(int i = 1; i <= 4; i++) {
            if (!str[i]) {
                for(int j = 1; j <= 4; j++) {
                    if (i == j) continue;
                    if (str[j]) continue;
                    str[i] = true;
                    long k = arr[j];
                    if (t[cnt] == '+') {
                        arr[j] += arr[i];
                    } else {
                        arr[j] *= arr[i];
                    }
                    dfs(cnt + 1);
                    arr[j] = k;
                    str[i] = false;
                }
            }
        }
    }

    @Override
    public void run() {


        arr[1] = cin.nextInt();arr[2] = cin.nextInt();arr[3] = cin.nextInt();arr[4] = cin.nextInt();
        t[1] = cin.next().charAt(0);t[2] = cin.next().charAt(0);t[3] = cin.next().charAt(0);
        dfs(1);
        cout.println(ans);
        cout.flush();
    }

}