Kalorona

Codeforces Round #534 (Div. 1) – 解题报告

A – Grid game

竖的横的上下分开填。

// CF1103A.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e5 + 200;

int n, ux, uy, dx, dy;
char str[MAX_N];

int main()
{
    scanf("%s", str + 1), n = strlen(str + 1);
    ux = 1, uy = 1, dx = 0, dy = 1;
    for (int i = 1; i <= n; i++)
        if (str[i] == '0')
            printf("%d %d\n", ux, uy), uy = uy % 4 + 1;
        else
            printf("%d %d\n", (dx >> 1) + 3, dy), dx = (dx + 1) % 4, dy = (dy + 1) % 4 + 1;
    return 0;
}

B – Game with modulo

一共有 60 次询问机会，我们可以分两个 30 次：第一个 30 次用来询问最高位前的一位，第二个 30 位用来补全我们对 \(a – 1\) 的猜测。

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牛客 CSP-S 提高组赛前集训营 1 – 解题报告

A – 仓鼠的石子游戏

诶嘿题。画了几张图发现只有\(1\)的情况先手必胜，所以考虑多轮交换先后手即可。

// A.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e3 + 200, table[8] = {0, 0, 1, 1, 1, 1, 1, 1};

int T, n, ai[MAX_N];

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", &ai[i]);
        bool flag = false;
        for (int i = 1; i <= n; i++)
            flag ^= (ai[i] == 1);
        printf(flag ? "rabbit\n" : "hamster\n");
    }
    return 0;
}

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「牛客 OI 周赛2 – 提高组」解题报告

A – 游戏

我大概乱搞一下：从最外围向内操作，计算操作次数再取模然后对应姓名即可。我不太了解为什么就过了。

// A.cpp
#include <bits/stdc++.h>

using namespace std;

const int MAX_N = 1e3 + 200;

char mp[MAX_N][MAX_N];
int n, m, T, delta[MAX_N][MAX_N], now[MAX_N][MAX_N], matrix[MAX_N][MAX_N];

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        memset(delta, 0, sizeof(delta)), memset(now, 0, sizeof(now));
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i++)
            scanf("%s", mp[i] + 1);
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= m; j++)
                if (mp[i][j] == 'R')
                    matrix[i][j] = 1;
                else if (mp[i][j] == 'G')
                    matrix[i][j] = 2;
                else
                    matrix[i][j] = 0;
        int cnt = 0;
        for (int i = n; i >= 1; i--)
        {
            delta[i][m + 1] += delta[i + 1][m + 1];
            for (int j = m; j >= 1; j--)
            {
                delta[i][j] += delta[i + 1][j];
                now[i][j] = now[i][j + 1] + delta[i][j + 1];
                int curt = (matrix[i][j] + now[i][j]) % 3;
                cnt += (3 - curt) % 3;
                now[i][j] += (3 - curt) % 3, delta[i][j + 1] += (3 - curt) % 3;
            }
        }
        printf("%s\n", cnt % 3 == 2 ? "dreagonm" : (cnt % 3 == 1 ? "fengxunling" : "BLUESKY007"));
    }
    return 0;
}

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NOIp 2016 解题报告

A – 玩具谜题

sb 题，随便搞。

// P1563.cpp
#include <iostream>

using namespace std;

int main()
{
    int N, M;
    cin >> N >> M;
    int dir[N];
    string occupations[N];
    int cmdKeys[M];
    int cmdPath[M];
    // I/O;
    for (int i = 0; i < N; i++)
        cin >> dir[i] >> occupations[i];
    for (int j = 0; j < M; j++)
        cin >> cmdKeys[j] >> cmdPath[j];
    // Process;
    int currentBias = 0;
    for (int i = 0; i < M; i++)
        if (cmdKeys[i] == 0)
            if (dir[currentBias % N] == 1)
                currentBias += cmdPath[i];
            else
            {
                currentBias -= cmdPath[i];
                if (currentBias < 0)
                    currentBias = N + currentBias;
            }
        else if (dir[currentBias % N] == 1)
        {
            currentBias -= cmdPath[i];
            if (currentBias < 0)
                currentBias = N + currentBias;
        }
        else
            currentBias += cmdPath[i];
    cout << occupations[currentBias % N];
    return 0;
}

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「Fortuna OJ」Aug 21 – Group A 解题报告

B – Maja

考虑超级暴力，设计状态\(dp[k][i][j]\)为走了\(k\)步处于\((i, j)\)的答案：

\[ dp[k][i][j] = mp[i][j] + \max \begin{cases} dp[k – 1][i – 1][j] \\ dp[k – 1][i + 1][j] \\ dp[k – 1][i][j + 1] \\ dp[k – 1][i][j – 1] \end{cases} \]

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