给三个数,a,b,k, 问a和b是不是能在k次(exactly) 下通过整除相等.

这个题情况比较多, 比如a和b互质, 比如k大于a和b. 反正就是交一次就对的都是大神.

这题思路是, 反正a和b一直和某数做整除, 最后就是1, 肯定相等啊, 所以只需要计算有多少质因子在a和b的和中.即可, 剩下就是各种corner cases.

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);  
    int t;
    cin >> t;
    while (t --)
    {
        auto gcd = [&](auto wtf, int a, int b) -> int {
        if(b == 0)
            return a;
        return wtf(wtf, b, a % b);
        }; 
        auto fac = [&](int n) -> int {
            int res = 0;
            for (size_t i = 2; i * i <= n; i++)
            {
                while (n % i == 0)
                {
                    res++;
                    n /= i;
                }
            }
            if (n > 1) // if n is a prime, but not 1
            {
                res++;
            }
            return res;
        };
        int a,b,k;
        cin >> a >> b >> k;
      //  cout << fac(a) << endl;
      //  cout << fac(b) << endl;
        if (k == 1)
        { 
            int g = gcd(gcd, a, b);
            if ((g == a || g == b) && a != b)
            {
                cout << "YES" << endl;
            }
            else
            {
                cout << "NO" << endl;
            }
        }
        else if (k > 1)
        {
            if (k > fac(a) + fac(b))
            {
                cout << "NO" << endl;
            }
            else
            {
                cout << "YES" << endl;
            }
        }
    } 
    
  
    
    return 0;
}

给一组数和l还有r, 求这组数中任意两个数的和在[L,R]中.

先排序, 然后求[0,R], 然后再求[0,L-1], 两个相减即可.

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    int t;
    cin >> t;
    while (t --)
    {
        int n,l,r;
        cin >> n >> l >> r;
        vector<int> v; 
        int k = n;
        while (k--)
        {
            int vv;
            cin >> vv;
            v.push_back(vv);
        }
        int res = 0;
        sort(v.begin(), v.end());
        auto count = [&](int r) -> long long{
            long long res = 0;
            int i = 0, j = n - 1;
            while(i < j) {
                while(i < j && v[i] + v[j] > r)
                {
                    j--;
                }
                res += (j - i);
                i++;
            }
            return res;
        };
        cout << count(r) - count(l - 1) << endl;
    }
    return 0;
}

给一个数组, 可以从两端取数字, 取了就没了, 求最少多少步可以取了最大和最小值.

这题他妈的真费时间, 一共3种取法 (答案给的是四种), 反正我三种做的, 三种是: 从左往右取, 从右往左和两侧取.

这题给了最大值是n, 最小值是1, 所以1/2两种都很好做, 就是第三个, 需要找下两个值的坐标, 然后减去两端就行了.

static class TaskA {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int maxIndex = -1;
            int minIndex = -1;
            int n = in.readInt();
            int[] ary = in.readIntArray(n);
            int left = 0, right = 0, both = 0;
            for (int i = 0; i < n; i++) {
                if (ary[i] == 1)
                    minIndex = i;
                else if (ary[i] == n)
                    maxIndex = i;
            }
            boolean one = false;
            boolean two = false;
            for (int i = 0; i < n; i++) {
                if (ary[i] == 1)
                    one = true;
                else if (ary[i] == n)
                    two = true;
                if (one && two) {
                    left = i + 1;
                    break;
                }
            }
            one = false;
            two = false;
            for (int i = n - 1; i >= 0; i--) {
                if (ary[i] == 1)
                    one = true;
                else if (ary[i] == n)
                    two = true;
                if (one && two) {
                    right = n - i;
                    break;
                }
            }
            both = Math.min(minIndex, maxIndex) + 1 + n - Math.max(minIndex, maxIndex);
            out.printLine(Math.min(Math.min(left, right), both));
        }
 
    }

给一个字符串s, 求字典序最小的字符串没有在s中出现.

这题就是bfs按个试..妈的, 比赛的时候想多了. 浪费时间

package readman;

import net.egork.io.InputReader;
import net.egork.io.OutputWriter;

import java.util.*;

public class TaskB {
    public void solve(int testNumber, InputReader in, OutputWriter out) {
        int n = in.readInt();
        String s = in.readString();
         List<String> list = new ArrayList<>();
         list.add("");
        while (true) {
            int size = list.size();
            for (int i = 0; i < size; i++){
                if (list.size() > 1 && list.get(i) == "")
                    continue;
                for (int j = 0; j < 26; j++) {
                    String ss = list.get(i) + (char) ('a' + j);
                     if (!s.contains(ss)){
                        out.printLine(ss);
                        return;
                    }
                     list.add(ss);
                }
            }
        }
    }
}

给一个数组, 问是否可以组成一个数组, 里面的元素包含其中任意两个元素相减的绝对值.

这题, 我上来以为是等差数列什么的呢…结果就是暴力求解..

    static class TaskA {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int n = in.readInt();
            int[] ary = in.readIntArray(n);
            Set<Integer> set = new HashSet<>();
            List<Integer> list = new ArrayList<Integer>();
            for (int a : ary) {
                set.add(a);
                list.add(a);
            }
            boolean add = true;
            while (set.size() <= 300 && add) {
                add = false;
                for (int i = 0; i < list.size(); i++) {
                    for (int j = i + 1; j < list.size(); j++) {
                        int nn = Math.abs(list.get(i) - list.get(j));
                        if (!set.contains(nn)) {
                            set.add(nn);
                            list.add(nn);
                            add = true;
                            break;
                        }
                    }
                }
            }
            if (set.size() > 300) {
                out.printLine("NO");
            } else {
                out.printLine("YES");
                out.printLine(list.size());
                for (int l : list) {
                    out.print(l + " ");
                }
                out.printLine();
            }
        }
 
    }