2019省赛训练八(2017 ECNA Regional Contest)

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2017 ECNA Regional Contest

题目

链接: https://pan.baidu.com/s/1EACO23Tlp83df59WkzP0mQ 提取码: uvrr

题解

Problem C DRM Messages

#include <bits/stdc++.h>

using namespace std;

int main()
{
    string s;
    cin >> s;
    string s1 = s.substr(0, s.length() / 2);
    string s2 = s.substr(s.length() / 2);
    int sum1 = 0, sum2 = 0;
    for (int i = 0; i < s.length() / 2; i++)
    {
        sum1 += s1[i] - 'A';
        sum2 += s2[i] - 'A';
    }
    for (int i = 0; i < s.length() / 2; i++)
    {
        s1[i] = (s1[i] - 'A' + sum1) % 26 + 'A';
        s2[i] = (s2[i] - 'A' + sum2) % 26 + 'A';
        s1[i] = (s1[i] - 'A' + s2[i] - 'A') % 26 + 'A';
    }
    cout << s1 << endl;
    return 0;
}

Problem D Game of Throwns

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n, m;
    stack<int> ope;
    cin >> n >> m;
    int pos = 0;
    while (m--)
    {
        string s;
        cin >> s;
        if (s == "undo")
        {
            cin >> s;
            int u = stoi(s);
            while (u-- && !ope.empty())
                ope.pop();
        }
        else
            ope.push(stoi(s));
    }
    while (!ope.empty())
    {
        pos += ope.top();
        ope.pop();
    }
    while (pos < 0)
        pos += n;
    cout << pos % n << endl;
    return 0;
}

Problem H Sheba’s Amoebas

#include <bits/stdc++.h>

using namespace std;

int a, b;
char m[101][101];
const int dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
const int dy[] = {1, -1, 0, 1, -1, 0, 1, -1};
struct cor
{
    int x, y;
};
bool vis[101][101];

bool is_inside(int i, int j)
{
    return i >= 1 && i <= a && j >= 1 && j <= b ? true : false;
}

bool BFS(int i, int j)
{
    bool flag = true;
    queue<cor> q;
    q.push(cor{i, j});
    while (!q.empty())
    {
        cor c = q.front();
        q.pop();
        vis[c.x][c.y] = true;
        int cnt = 0;
        for (int k = 0; k < 8; k++)
            if (is_inside(c.x + dx[k], c.y + dy[k] && m[c.x + dx[k]][c.y + dy[k]] == '#'))
            {
                cnt++;
                if (!vis[c.x + dx[k]][c.y + dy[k]])
                    q.push(cor{c.x + dx[k], c.y + dy[k]});
            }
        if (cnt != 2)
            flag = false;
    }
    return flag ? true : false;
}

int main()
{
    scanf("%d%d", &a, &b);
    getchar();
    for (int i = 1; i <= a; i++)
    {
        for (int j = 1; j <= b; j++)
            m[i][j] = getchar();
        getchar();
    }
    int ans = 0;
    for (int i = 1; i <= a; i++)
        for (int j = 1; j <= b; j++)
            if (!vis[i][j] && m[i][j] == '#')
                ans += BFS(i, j);
    printf("%d\n", ans);
    return 0;
}