登录 注册

SDWU 2022 ACM Individual Training Match 1st 题解

本贴最后更新于 974 天前，其中的信息可能已经东海扬尘

比赛：SDWU 2022 ACM Individual Training Match 1st(已结束)
最终比赛榜单：SDWU 2022 ACM Individual Training Match 1st 比赛榜单
比赛题目练习：SDWU 2022 ACM Individual Training Match 1st 训练

每场比赛结束后都会创建该比赛的相关练习，点击上方《比赛题目练习》的链接后即可进入，请使用之前下发的个人账号进行练习！直接点击题解中每题的标题也可直达练习页面！

请注意！！！提交时编译语言请选择：GNU G++17 7.3.0

G++ 同时兼容 C++ 与 C 语言，在正式比赛中也尽量选择高版本编译器

因远程评测题目网络问题，提交后可能有较长时间的 Pending 时间，请耐心等待，若出现 Submitted Failed 问题，请点击刷新符号重新进行提交！

image.png

A - All in!

#include <bits/stdc++.h>
using namespace std;
int main() {
    puts("All in!");
    return 0;
}

B - Boboge and Tall Building

#include <bits/stdc++.h>
using namespace std;
int T, n, m, k;
int main() {
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d%d%d", &n, &m, &k);
        double h = k * 1.0 / m;
        printf("%.10lf\n", h * (n - 1) * 1.0);
    }
    return 0;
}

C - Constructive Problem

打表并找规律

//打表
#include <bits/stdc++.h>
using namespace std;
const int N = 10;
int n;
int st[N];
int main() {
    scanf("%d", &n);
    function<void(int, int)> dfs = [&](int d, int sum) {
        if (sum > n) return;
        if (d == n) {
            if (sum != n)   return;
            vector<int> cnt(20); 
            for (int i = 0; i < n; i ++ )
                cnt[st[i]] ++;
            bool flag = 1;
            for (int i = 0; i < n; i ++ ) {
                if (cnt[i] != st[i]) {
                    flag = 0;
                    break;
                }
            }
            if (flag) {
                for (int i = 0; i < n; i ++ )
                    printf("%d ", st[i]);
                puts("");
            }
            return;
        }
        for (int i = 0; i < n; i ++ ) {
            st[d] = i;
            dfs(d + 1, sum + i);
        }
    };
    dfs(0, 0);
    return 0;
}

//提交代码
#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
    scanf("%d", &n);
    if (n < 4 || n == 6)    puts("-1");
    else if (n == 4)    puts("1 2 1 0");
    else if (n == 5)    puts("2 1 2 0 0");
    else {
        printf("%d 2 1 ", n - 4);
        for (int i = 1; i <= n - 7; i ++ )
            cout << 0 << " ";
        cout << 1;
        cout << " 0 0 0" << endl;
    }
    return 0;
}

D - Diseased String

#include <bits/stdc++.h>
using namespace std;
int T, n, m, ans;
char s[110];
int main() {
    scanf("%d", &T);
    while (T -- ) {
        ans = 0;
        scanf("%d", &n);
        scanf("%s", s + 1);
        for (int i = 1; i <= n; ) {
            if (s[i] != 'y') {
                i ++;
                continue;
            } 
            int j = i + 1;
            while (j  <= n && s[j] == 'b')  j ++;
            if (j - i >= 3) ans += j - i - 2;
            i = j;
        }
        printf("%d\n", ans);
    }
    return 0;
}

E - Equality

贪心
特判全部相等以及 m=1 的情况
首先将数组转换为 01 数组，1 表示当前数为最小值，0 表示当前数不为最小值
找到第一个 1 的位置，因为要使得前面的所有数都为 1，只能通过第一个 1 的位置向前拓展
之后从消除完第一个 1 的位置的前面所有 0 所得到的最远的起始位置
通过前缀和来快速判断一个区间中是否含有 1
如果当前区间含有 1，那么一次可以向后拓展 m 个位置，否则只能拓展 m−1 个位置

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int T, n, m, minn;
int a[N], s[N];
int main() {
    scanf("%d", &T);
    while (T -- ) {
        minn = 0x3f3f3f3f;
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; i ++ ) {
            scanf("%d", &a[i]);
            minn = min(minn, a[i]);
        }
        bool flag = 1;
        for (int i = 1; i <= n; i ++ )
            if (a[i] != minn)
                flag = 0;
        if (flag)   puts("0");
        else if (m == 1)    puts("-1");
        else {
            int pos = 0;
            int cnt = 0;
            memset(s, 0, sizeof s);
            for (int i = 1; i <= n; i ++ ) {
                if (a[i] == minn) {
                    pos = i;
                    break;
                }
            }
            int k = (int)ceil((pos - 1) * 1.0 / (m - 1));
            cnt += k;
            int last = max(pos, (m - 1) * k + 1);
            for (int i = 1; i <= last; i ++ )   a[i] = minn;
            for (int i = 1; i <= n; i ++ ) 
                s[i] = s[i - 1] + (a[i] == minn ? 1 : 0);
            for (int i = last; i <= n; ) {
                if (a[i] == minn) {
                    i ++;
                    continue;
                }
                else {
                    cnt ++;
                    if (s[min(n, i + m - 1)] - s[i - 1] > 0) i = i + m;
                    else i = i + m - 1;
                }
            } 
            printf("%d\n", cnt);
        }
    }
    return 0;
}

F - Future Vision

BFS

#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int T, n, m;
int d[N][N];
bool st[N][N];
char s[N], g[N][N];
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
struct node {
    int x, y;
    int step;
};
int main() {
    scanf("%d", &T);
    while (T -- ) {
        memset(st, 0, sizeof st);
        memset(d, 0x3f, sizeof d);
        scanf("%d%d", &n, &m);
        int x, y;
        for (int i = 1; i <= n; i ++ ) {
            scanf("%s", s + 1);
            for (int j = 1; j <= m; j ++ ) {
                g[i][j] = s[j];
                if (g[i][j] == 'H')
                    x = i, y = j;
            }
        }
        function<void()> bfs = [&]() {
            queue<node> q;
            q.push({x, y, 0});
            d[x][y] = 0;
            st[x][y] = true;
            while (q.size()) {
                auto t = q.front();
                q.pop();
                for (int i = 0; i < 4; i ++ ) {
                    int tx = t.x + dx[i];
                    int ty = t.y + dy[i];
                    if (tx < 1 || tx > n || ty < 1 || ty > m || st[tx][ty]|| g[tx][ty] == '#') continue;
                    d[tx][ty] = t.step + 1;
                    st[tx][ty] = 1;
                    q.push({tx, ty, t.step + 1});
                }
            }
        };
        bfs();
        int time, ans;
        bool flag = 0;
        scanf("%d", &time);
        for (int t = 0; t < time; t ++ ) {
            int a, b;
            scanf("%d%d", &a, &b);
            if (g[a][b] == '#') continue;
            if (d[a][b] <= t && !flag) {
                ans = t;
                flag = 1;
            }
        }
        if (flag)   printf("YES %d\n", ans);
        else puts("NO");
    }
    return 0;
}

G - Generate 7 Colors

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 10;
int T, n;
int a[N];
signed main() {
    scanf("%lld", &T);
    while (T -- ) {
        for (int i = 0; i < 7; i ++ )
            scanf("%lld", &a[i]);
        bool flag = 1;
        for (int i = 1; i < 7; i ++ )
            if (a[i] > a[i - 1])
                flag = 0;
        if (flag == 0)  {printf("-1\n"); continue;}
        int cnt = 0;
        for (int i = 6; i >= 0; i -- ) {
            if (a[i] == 0)  continue;
            if (i != 6) cnt += a[i];
            else cnt ++;
            for (int j = 0; j <= i; j ++ )
                a[j] -= a[i];
            if (i == 6) 
                for (int j = 0; j <= i; j ++)
                    if (a[j])   
                        a[j] --;  
            if (a[0] == 0)  break;
        }
        printf("%lld\n", cnt);
    }
    return 0;
}

H - Hile and Subsequences' MEX

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MOD = 998244353;
int T, n;
int qmi(int m, int k, int p)
{
    int res = 1 % p, t = m;
    while (k) {
        if (k & 1) res = res * t % p;
        t = t * t % p;
        k >>= 1;
    }
    return res;
}
signed main() {
    scanf("%lld", &T);
    while (T -- ) {
        scanf("%lld", &n);
        printf("%d\n", qmi(2, n, MOD) - 1);
    }
    return 0;
}

I - If I Catch You

该题无题解(水平有限)，如果您做出了这道题，欢迎投稿，我会及时更新文档~

J - Jiubei and Codeforces

模拟

#include <bits/stdc++.h>
using namespace std;
int T, n, k, x;
struct node {
    int l, r;
    string s;
}p[20];
int main() {
    p[1].l = 0xcfcfcfcf, p[1].r = 1199, p[1].s = "Newbie";
    p[2].l = 1200, p[2].r = 1399, p[2].s = "Pupil";
    p[3].l = 1400, p[3].r = 1599, p[3].s = "Specialist";
    p[4].l = 1600, p[4].r = 1899, p[4].s = "Expert";
    p[5].l = 1900, p[5].r = 2099, p[5].s = "Candidate master";
    p[6].l = 2100, p[6].r = 2299, p[6].s = "Master";
    p[7].l = 2300, p[7].r = 2399, p[7].s = "International master";
    p[8].l = 2400, p[8].r = 2599, p[8].s = "Grandmaster";
    p[9].l = 2600, p[9].r = 2999, p[9].s = "International grandmaster";
    p[10].l = 3000, p[10].r = 0x3f3f3f3f, p[10].s = "Legendary grandmaster";
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d%d", &n, &k);
        string s1, s2;
        for (int i = 1; i <= n; i ++ ) {
            int rate = k;
            scanf("%d", &x);
            k += x;
            for (int i = 1; i <= 10; i ++ ) {
                if (p[i].l <= k && p[i].r >= k) s1 = p[i].s;
                if (p[i].l <= rate && p[i].r >= rate) s2 = p[i].s;
            }
            if (s1 != s2) cout << s2 << " -> " << s1 << endl;
        }
        cout << s1 << endl;
    }
    return 0;
}

K - Klee and Bomb

该题无题解(水平有限)，如果您做出了这道题，欢迎投稿，我会及时更新文档~

L - Lexicographic Order

如果最后一位是 a，直接删除最后一位输出 s
如果最后一位不是 a，最后一位减 1，用 z 填充剩下的位数直到 m

#include <bits/stdc++.h>
using namespace std;
int n, m;
string s;
int main() {
    cin >> n >> m >> s;
    if (s[s.size() - 1] == 'a')
        s = s.substr(0, s.size() - 1);
    else {
        char c = s[s.size() - 1];
        c --;
        s = s.substr(0, s.size() - 1);
        s += c;
        while (s.size() < m)
            s += "z";
    }
    cout << s << endl;
    return 0;
}
2 操作
Shun2002 在 2022-04-27 23:03:15 更新了该帖
Shun2002 在 2022-04-27 23:02:37 更新了该帖
香港 位置
2 操作
75 7 8 20 6 7 27

相关帖子

回帖
SDWU 2022 ACM Individual Training Match 1st 题解

欢迎来到这里！

我们正在构建一个小众社区，大家在这里相互信任，以平等 • 自由 • 奔放的价值观进行分享交流。最终，希望大家能够找到与自己志同道合的伙伴，共同成长。

注册 关于
请输入回帖内容 ...
Shun2002
努力 坚持 香港
回帖
1
 帖子
6
 积分
358