JXUST_NC - ACM工作室20级选拔赛题解

A - RioTian学长的星际航线

并查集板子

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
int n, m;  //星球数和星际航线数M。
int a, b, father[maxn];
int find(int x) {
    int r = x;
    while (r != father[r]) r = father[r];
    int i = x, j;
    while (father[i] != r) {
        j = father[i];
        father[i] = r;
        i = j;
    }
    return r;
}
int main() {
    while (cin >> n && n) {
        cin >> m;
        //初始化。
        for (int i = 1; i <= n; i++) father[i] = i;
        while (m--) {
            cin >> a >> b;
            a = find(a), b = find(b);
            if (a != b) father[a] = b;
        }
        int cnt = 0;
        for (int i = 1; i <= n; i++) {
            if (father[i] == i) cnt++;
        }
        cout << cnt - 1 << endl;
    }
    return 0;
}

B - 江理大家庭

根据题意不断模拟。

#include <bits/stdc++.h>

using namespace std;

string s;
bool judge1(char a) {
    if (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u') return true;
    else
        return false;
}
bool judge2(string s) {
    bool flag1 = false, flag2 = true, flag3 = true;
    //若为1说明前者为元音，若为-1说明还没有确定，若为0说明前者为辅音。
    int flag = -1;
    int len = s.size();
    int cnt = 1;  //统计连续的原因。
    char pre = '0';
    for (int i = 0; i < len; i++) {
        if (judge1(s[i])) flag1 = true;
        if (pre != '0') {
            //连续字母的判断。
            if (pre == s[i]) {
                if (pre != 'e' && pre != 'o') flag3 = false;
            } else
                pre = s[i];
        }
        if (flag != -1) {
            if (flag) {
                //说明为元音，那么判断后者是否为元音。
                if (judge1(s[i])) cnt++;
                else {
                    //为辅音就改变。
                    cnt = 1;
                    flag = 0;
                }
            } else {
                if (!judge1(s[i])) cnt++;
                else {
                    cnt = 1;
                    flag = 1;
                }
            }
            if (cnt == 3) flag2 = false;
        }
        if (pre == '0') pre = s[i];
        if (flag == -1) {
            if (judge1(s[i])) flag = 1;
            else
                flag = 0;
        }
    }
    if (flag1 && flag2 && flag3) return true;
    else
        return false;
}
int main() {
    while (cin >> s) {
        if (s == "end") break;
        cout << "<" << s << "> ";
        cout << (judge2(s) ? "is acceptable.\n" : "is not acceptable.\n");
    }
    return 0;
}

C - 粗心的RioTian学长

数组绝对值排序

#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n, a[maxn];
bool cmp(int a, int b) { return abs(a) > abs(b); }
int main() {
    while (cin >> n) {
        for (int i = 0; i < n; i++) cin >> a[i];
        sort(a, a + n, cmp);
        for (int i = 0; i < n; i++) {
            cout << a[i];
            i == n - 1 ? cout << endl : cout << " ";
        }
    }
    return 0;
}

D - 女装大佬RioTian

利用双指针寻找

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
int t, n, s, min_len;
int a[maxn];
int main() {
    cin >> t;
    while (t--) {
        cin >> n >> s;
        for (int i = 0; i < n; i++) cin >> a[i];
        min_len = maxn;
        int sum = 0, cur = 0, last = 0;  //双指针。
        while (cur < n) {
            sum += a[cur];
            while (sum >= s) {
                min_len = min(min_len, cur - last + 1);
                sum -= a[last++];
            }
            cur++;
        }
        if (min_len == maxn) {
            cout << "0" << endl;
        } else {
            cout << min_len << endl;
        }
    }
    return 0;
}

E - RioTian学长的数组

简单贪心思想

字典序最小，首先让最前面的最小，最后面最大

void solve() {
    int n, k;
    cin >> n >> k;
    int a[n + 1];
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int i = 1, j = n;
    while (true) {
        if (i == j || k == 0) break;
        if (a[i] >= 1) a[i] -= 1, a[j] += 1, k--;
        else
            i++;
    }
    for (int i = 1; i <= n; ++i) cout << a[i] << " ";
    cout << "\n";
}

F - RioTian学长爱照相

要使 (au + av) / 2 为整数的上镜对越多，那么简单来说，奇数和奇数成对，偶数与偶数组队，这样上镜对一定最大

void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    vector<int> odd, even;
    for (int &x : a) {
        cin >> x;
        if (x & 1) odd.push_back(x);
        else
            even.push_back(x);
    }
    for (int x : odd) cout << x << " ";
    for (int x : even) cout << x << " ";
    cout << '\n';
}

G - RioTian学长与矩阵

概括一下题意：在一个被限定了宽度的盒子中给一些长度为 $2^x$ 高度为 1 的长方块，请问怎么排放才能使最终高度最小

思路：先利用 map 存储各个长度的值，然后二分找到在该行中最大的一块然后填充。

void solve() {
    int n, w;
    cin >> n >> w;
    map<int, int> mp;
    for (int i = 0, x; i < n; ++i) {
        cin >> x;
        mp[x]++;
    }
    int ans = 1, cur = w;
    while (mp.size()) {
        if (mp.begin()->first > cur) {
            ++ans, cur = w;
        }
        auto it = prev(mp.upper_bound(cur));
        assert(it->first <= cur);
        cur -= it->first;
        if (--(it->second) == 0) mp.erase(it);
    }
    cout << ans << "\n";
}

H - 字符串的最小开销

这是一道比较难的字符串思维题

总而言之就是让连续的两个ij 出现尽可能少的次数，尽可能平均，所以直接让ababc....akbcbdbe...bkc这样过去，然后最后到(k-1)k 又连到a，刚好所有连续两个出现了一次，这样还不够的话继续循环下去就行了

void solve() {
    int n, k;
    cin >> n >> k;
    string s;
    for (int i = 0; i < k; ++i) {
        s += 'a' + i;
        for (int j = i + 1; j < k; ++j) s += 'a' + i, s += 'a' + j;
    }
    assert(s.size() == k * k);
    for (int i = 0; i < n; ++i) cout << s[i % s.size()];
}

I - RioTian学长爱玩消消乐

首先给数组排序，如果存在相邻差值大于2的则输出 NO

#include <bits/stdc++.h>
using namespace std;
int main() {
    int t, n, a[1100];
    cin >> t;
    while (t--) {
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        sort(a, a + n);
        int flag = 0;
        for (int i = 1; i < n && !f; i++) {
            if (abs(a[i] - a[i - 1]) >= 2)
                flag = 1;
        }
        if (flag)
            cout << "NO" << endl;
        else
            cout << "YES" << endl;
    }
}

J - Hello，World！

签到题

cout << "hello world";