UCF Local Programming Contest 2016

题目链接：

https://www.jisuanke.com/contest/7194?view=challenges

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                    だからよ...止まるじゃねえぞ
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A. Majestic 10

题意：

给你三个数判断有几个大于等于10，0个输出zilch，一个输出double，两个输出double-double，三个输出triple-double

思路：

无

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<double, int> PII;
const int N = 100000 + 5;
const double eps = 1e-9;

int main() {
    int n;
    cin >> n;
    while (n--) {
        int ans = 0;
        int a, b, c;
        cin >> a >> b >> c;
        if (a >= 10)
            ans++;
        if (b >= 10)
            ans++;
        if (c >= 10)
            ans++;
        cout << a << " " << b << " " << c << endl;
        if (ans == 0)
            cout << "zilch" << endl;
        else if (ans == 1)
            cout << "double" << endl;
        else if (ans == 2)
            cout << "double-double" << endl;
        else if (ans == 3)
            cout << "triple-double" << endl;
        if (n)
            cout << endl;
    }
    return 0;
}

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B. Phoneme Palindromes

题意：

博主太懒了，什么都没有写

思路：

博主太懒了，什么都没有写

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<double, int> PII;
const int N = 500 + 5;
const double eps = 1e-9;

set<char> v[N];

bool check(char a, char b) {
    for (auto i:v[a])
        if (i == b)
            return true;
    return false;
}

int main() {
    int t;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        cout << "Test case #" << i << ":" << endl;
        int p;
        cin >> p;
        for (int j = 1; j <= p; j++) {
            getchar();
            char a, b;
            cin >> a >> b;
            v[a].insert(b);
            v[b].insert(a);
        }
        int n;
        cin >> n;
        string s;
        while (n--) {
            cin >> s;
            int l = 0, r = s.length() - 1;
            bool flag = true;
            while (l <= r) {
                if (!(s[l] == s[r] || check(s[l], s[r]))) {
                    flag = false;
                    break;
                }
                l++;
                r--;
            }
            cout << s << " ";
            if (flag)
                cout << "YES" << endl;
            else
                cout << "NO" << endl;
        }
        for (auto &j : v)
            j.clear();
        if (i != t)
            cout << endl;
    }
    return 0;
}

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D. Wildest Dreams

题意：

 女儿在奇数时间段里在车上，偶数时间段不在车上

 一旦在车上就就要调到女儿喜欢的歌曲上

 女儿在车不管到了哪首歌，都要立即跳到女儿喜欢的歌，不在车就按列表循环播放，求爸爸听了女儿喜欢的歌多少秒

思路：

忘了，我想想

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<double, int> PII;
const int N = 500 + 5;
const double eps = 1e-9;

int a[N];

int main() {
    int t;
    cin >> t;
    for (int o = 1; o <= t; o++) {
        cout << "CD #" << o << ":" << endl;
        int n, p;
        cin >> n >> p;
        p--;
        for (int i = 0; i < n; i++)
            cin >> a[i];
        int x;
        cin >> x;
        while (x--) {
            ll ans = 0;
            int m;
            cin >> m;
            int now = 0, timer = 0;
            for (int i = 1; i <= m; i++) {
                int pp;
                cin >> pp;
                if (i % 2) {
                    now = p;
                    timer = pp % a[p];
                    ans += pp;
                } else {
                    if (timer == 0)
                        now = (now + 1) % n;
                    while (pp) {
                        if (now == p) {
                            if (a[now] - timer >= pp) {
                                timer += pp;
                                ans += pp;
                                pp = 0;
                            } else {
                                ans += a[now] - timer;
                                pp -= a[now] - timer;
                                timer = 0;
                                now = (now + 1) % n;
                            }
                        } else {
                            if (a[now] - timer >= pp) {
                                timer += pp;
                                pp = 0;
                            } else {
                                pp -= a[now] - timer;
                                timer = 0;
                                now = (now + 1) % n;
                            }
                        }
                    }
                }
            }
            cout << ans << endl;

        }

        if (o != t)
            cout << endl;
    }
    return 0;
}

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E. Loopy Word Search

题意：

给你一个字母表，求给出的单词出现在字母表里的位置和方向

思路：

按题意来就行了，简单的暴力

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<double, int> PII;
const int N = 500 + 5;
const double eps = 1e-9;

int r, c;
string s[N];
int xi[4] = {0, 1, 0, -1};
int yi[4] = {1, 0, -1, 0};

bool dfs(int len, int fang, int x, int y, string now) {
    int nowx = x;
    int nowy = y;
    for (int k = 1; k <= len; k++) {
        nowx = (nowx + xi[fang] + r) % r;
        nowy = (nowy + yi[fang] + c) % c;
        if (s[nowx][nowy] != now[k])
            return false;
    }
    return true;
}

int main() {
    int t;
    cin >> t;
    for (int k = 1; k <= t; k++) {
        cout << "Word search puzzle #" << k << ":" << endl;
        cin >> r >> c;
        for (int i = 0; i < r; i++)
            cin >> s[i];
        int n;
        cin >> n;
        while (n--) {
            string m;
            cin >> m;
            int len = m.length() - 1;
            bool flag = false;
            int x = 0, y = 0, front = 0;
            for (int i = 0; i < r; i++) {
                for (int j = 0; j < c; j++) {
                    if (s[i][j] == m[0])
                        for (int fang = 0; fang < 4; fang++) {
                            if (dfs(len, fang, i, j, m)) {
                                front = fang;
                                x = i;
                                y = j;
                                flag = true;
                                break;
                            }
                        }
                    if (flag)
                        break;
                }
                if (flag)
                    break;
            }
            if (front == 0)
                cout << "R" << " ";
            else if (front == 1)
                cout << "D" << " ";
            else if (front == 2)
                cout << "L" << " ";
            else if (front == 3)
                cout << "U" << " ";
            cout << x + 1 << " " << y + 1 << " " << m << endl;
        }
        if (k != t)
            cout << endl;
    }
    return 0;
}

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