一、4503. 数对数量

1. 题目描述

2. 思路分析

签到题,Just brute force the pairs of integers for the three conditions(x, y)的个数即可.

3. 代码实现

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int a, b, n;
    int res = 0;
    cin >> a >> b >> n;
    
    for (int i = 0; i <= a; i ++ ) 
    {
        for (int j = 0; j <= b; j ++ )
        if (i + j == n) res ++;
    }
    cout << res << endl;
    
    return 0;
}

二、4504. 字符串消除

1. 题目描述

2. 思路分析

1. Use the stack for maintenance,First push the first character onto the stack,依次枚举每个字符
   • If the stack is not empty and the character is the same as the top element of the stack,则栈顶元素出栈,并且ans++;
   • Otherwise push the character onto the stack;
2. 如果ans为奇数,则先手必赢;如果ans为偶数,则先手必输.

3. 代码实现

#include <bits/stdc++.h>
#include <cstring>

using namespace std;

const int N = 1e5 + 10;

char a[N];
stack<char> stk;

int main()
{
    int res = 0;
    scanf("%s", a);

    stk.push(a[0]);
    for (int i = 1; i < strlen(a); i ++ )
    {
        if (stk.size() > 0 && a[i] == stk.top())
        {
            stk.pop();
            res ++;
        }
        else
        {
            stk.push(a[i]);
        }
    }
    if (res % 2 == 0) cout << "No" << endl;
    else cout << "Yes" << endl;

    return 0;
}

三、4505. 最大子集

1. 题目描述

2. 思路分析

提示：The maximum possible size of a collection is 3,The smallest size will not be less than1,The case of three numbers is an arithmetic progression.

1. First put all the numbers into the hash set;
2. 枚举最小值,以及公差（最多30种公差）,找到长度3the longest sequence within ;
3. 剪枝：找到长度为3The sequence ends the algorithm directly.

3. 代码实现

#include <bits/stdc++.h>

using namespace std;

const int N = 200010, M = 1999997, INF = 0x3f3f3f3f;

int n;
int q[N], h[M];

int find(int x)
{
    int t = (x % M + M) % M;
    while (h[t] != INF && h[t] != x)
        if ( ++ t == M)
            t = 0;
    return t;
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i < n; i ++ ) scanf("%d", &q[i]);
    sort(q, q + n);

    memset(h, 0x3f, sizeof h);

    int res[3], s[3];
    int rt = 0, st = 0;
    for (int i = 0; i < n; i ++ )
    {
        for (int j = 0; j <= 30; j ++ )
        {
            int d = 1 << j;
            s[0] = q[i], st = 1;
            for (int k = 1; k <= 2; k ++ )
            {
                int x = q[i] - d * k;
                if (h[find(x)] == INF) break;
                s[st ++ ] = x;
            }
            if (rt < st)
            {
                rt = st;
                memcpy(res, s, sizeof s);
                if (rt == 3) break;
            }
        }
        if (rt == 3) break;
        h[find(q[i])] = q[i];
    }

    printf("%d\n", rt);
    for (int i = 0; i < rt; i ++ )
        printf("%d ", res[i]);

    return 0;
}

四、周赛总结

This week onlyac了前两道,The third question is too complicated to think aboutac出来,Fight for the next weekak.

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