Educational Codeforces Round 135 (Rated for Div. 2)

A

题意：
给定n个颜色的各自的数量，每次可以使用两个不同的颜料，问最后可能剩下哪种颜料，输出任意一个即可。

思路：

直接输出个数最多的那个颜料即可

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define lb(x) x &(-x)
#define endl "\n"
typedef pair<int, int> PII;
const int N = 1010, INF = 1e9;
int n, m, to;
struct st
{
    int a, b;
} s[N];

bool cmp(st a, st b)
{
    return a.a < b.a;
}

void solve()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> s[i].a;
        s[i].b = i;
    }
    sort(s + 1, s + 1 + n, cmp);
    cout << s[n].b << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int w = 1;
    cin >> w;
    while (w--)
        solve();
}

 

B

思路：

n为偶数，前 n-2 个数倒序输出，最后输出 n-1,n

n为奇数，先输出n-1,再输出前n-3个数，最后输出n-2,n

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define lb(x) x &(-x)
#define endl "\n"
typedef pair<int, int> PII;
const int N = 1010, INF = 1e9;
int n, m, to;
struct st
{
    int a, b;
} s[N];

bool cmp(st a, st b)
{
    return a.a < b.a;
}

void solve()
{
    cin >> n;
    if (n == 1)
    {
        cout << "1\n";
        return;
    }
    if (n == 3)
    {
        cout << "2 1 3\n";
        return;
    }
    if (n & 1)
    {
        cout << n - 1 << " ";
        for (int i = n - 3; i >= 1; i--)
            cout << i << " ";
        cout << n - 2 << " " << n << endl;
    }
    else
    {
        for (int i = n - 2; i >= 1; i--)
            cout << i << " ";
        cout << n - 1 << " " << n << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int w = 1;
    cin >> w;
    while (w--)
        solve();
}

 

 

C

思路：

先把所以相同的数消掉，再把大于9的数都进行一次操作，记录操作数，再消去所以相同的数，加上除了1之外所以数的个数即为答案

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define lb(x) x &(-x)
#define endl "\n"
typedef pair<int, int> PII;
const int N = 1010, INF = 1e9;
int n, m, to;
struct st
{
    int a, b;
} s[N];

void solve()
{
    cin >> n;
    map<int, int> ma, mb;
    for (int i = 1; i <= n; i++)
    {
        int a;
        cin >> a;
        ma[a]++;
    }
    int cnt = 0;
    for (int i = 1; i <= n; i++)
    {
        int a;
        cin >> a;
        if (ma[a] > 0)
            ma[a]--;
        else
        {
            if (a > 9)
            {
                cnt++;
                int t = log10(a) + 1;
                ma[t]--;
            }
            else
            {
                ma[a]--;
            }
        }
    }
    for (auto x : ma)
    {
        if (x.first > 9)
        {
            cnt += x.second;
            ma[log10(x.first) + 1] += x.second;
        }
    }
    for (int i = 2; i <= 9; i++)
    {
        cnt += abs(ma[i]);
    }
    cout << cnt << endl;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int w = 1;
    cin >> w;
    while (w--)
        solve();
}

 

 

D

题意：
给定一个字符串，Alice和Bob进行游戏，两个人各拥有一个初始为空的字符串，每人每次可以从字符串首或尾取一个字符放入自己字符串的头部，问再字符串取完后是Alice胜还是Bob胜还是平局。

分析：
区间dp，显然Bob不可能取胜。

那么对于区间[l,r]我们进行分类讨论.我们首先对整个dp数组初始化为0,都是平局.在进行分情况讨论:

如何使得dp[l][r]为 1呢

假设Alice取s[l].

1. 如果Bod取s[r]

　　 若dp[l+1,r-1]为1，结束，否则必须s[l]<s[r]

2.如果Bob取s[l+1]

　　若dp[l+2,r]为1，结束，否则必须s[l]<s[l+1]

什么两种情况必须同时满足

反之Alice取s[r],可以类比易得

 

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define lb(x) x &(-x)
#define endl "\n"
typedef pair<int, int> PII;
const int N = 1010, INF = 1e9;
int n, m;
void solve()
{
    string s;
    cin >> s;
    n=s.length();
    vector<vector<int>> dp(n, vector<int>(n));
    for (int len = 2; len <= n; len += 2)
        for (int i = 0; i < n; i++)
        {
            int j = i + len - 1;
            if (j >= n)
                break;
            if (len == 2)
            {
                dp[i][j] = (s[i] != s[j]);
                continue;
            }
            if ((dp[i + 1][j - 1] || s[i] < s[j]) && (dp[i + 2][j] || s[i] < s[i + 1]))
                dp[i][j] = 1;
            if ((dp[i + 1][j - 1] || s[j] < s[i]) && (dp[i][j - 2] || s[j] < s[j - 1]))
                dp[i][j] = 1;
        }
    cout << (dp[0][n - 1] ? "Alice" : "Draw") << endl;
}
signed main()
{
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int t = 1;
    cin >> t;
    while (t--)
        solve();
    return 0;
}