Codeforces Round #732 (Div. 2) A ~ D 个人题解记录

比赛链接：Here

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1546A - AquaMoon and Two Arrays

选定两个数组元素执行以下操作：

• $a_i,a_j(1\le i,j\le n)$ 一个 +1 另一个 -1，

  前提是两个数都要结果非负

请问在执行若干次后使得数组 $a$ 等于 数组 $b$

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先统计两个数组总和，只有和相同可以，否则输出 -1

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int _; for (cin >> _; _--;) {
        int n; cin >> n;
        int suma = 0, sumb = 0;
        vector<int>a(n), b(n);
        for (int &x : a)cin >> x, suma += x;
        for (int &x : b)cin >> x, sumb += x;
        if (suma != sumb) {
            cout << "-1\n";
            continue;
        }
        int cnt = 0;
        vector<pair<int, int>>ans;
        for (int i = 0; i < n; ++i)
            if (a[i] > b[i])ans.push_back({i, a[i] - b[i]}), cnt += (a[i] - b[i]);

        cout << cnt << "\n";
        if (cnt == 0)continue;

        int j = 0;
        for (int i = 0; i < n; ++i) {
            while (a[i] < b[i]) {
                if (ans[j].second > 0) {
                    a[i]++;
                    cout << ans[j].first + 1 << " " << i + 1 << "\n";
                    ans[j].second--;
                } else j++;
            }
        }
    }
}

1546B - AquaMoon and Stolen String

题意都能看懂就不写了...

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在每一列中，只会有一个元素出现奇数次，

只需要 map 存第 $i$ 位的值即可

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int _; for (cin >> _; _--;) {
        int n, m;
        cin >> n >> m;
        vector<ll>v(m, 0);
        string s;
        for (int i = 0; i < n; ++i) {
            cin >> s;
            for (int j = 0 ; j < m; ++j)
                v[j] += (s[j] - 'a');
        }

        for (int i = 0; i < n - 1; ++i) {
            cin >> s;
            for (int j = 0; j < m; ++j)
                v[j] -= (s[j] - 'a');
        }
        for (int i = 0; i < m; ++i) cout << char(v[i] + 'a');
        cout << '\n';
    }
}

1546C - AquaMoon and Strange Sort

对于每一个元素，肯定只能移动偶数距离

所以对于同一元素需要统计它们有多少个在奇数和偶数位置

原数组 $a$ ，排序后数组 $b$

对于每一个元素，如果它在 $a$ 中的奇数位置次数不同于在 $b$ 中奇数位置出现次数（偶数同理）则输出 NO，否则输出 YES

必须吐槽自己，大晚上写的代码不仔细，wa在 #38数据

const int N = 1e6 + 10;
int a[N], b[N], t[N][2];
int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    int _; for (cin >> _; _--;) {
        int n; cin >> n;
        memset(t, 0, sizeof(t));

        for (int i = 1; i <= n; ++i)cin >> a[i], b[i] = a[i];
        sort(b + 1, b + 1 + n);
        for (int i = 1; i <= n; ++i) {
            t[a[i]][i & 1]++;
            t[b[i]][i & 1]--;
        }

        bool f = true;
        for (int i = 1; f and  i <= 100000; ++i)
            if (t[i][0] || t[i][1]) f = false;
        cout << (f ? "YES\n" : "NO\n");
    }
}

1546D - AquaMoon and Chess

看完题感觉是像某种组合数学题，但没思路

先贴一下dalao代码

#define ll long long
#define int int64_t

constexpr int N = 1e5 + 5;
constexpr int INF = 1e9 + 5;
constexpr int mod = 998244353;

int n, fac[N], invfac[N];
string s;

int carp(int x, int y) { return x * y % mod;}

int binpow(int x, int y) {
    int res = 1;
    for (; y; y >>= 1, x = carp(x, x))
        if (y & 1) res = carp(res, x);
    return res;
}

int inv(int x) { return binpow(x, mod - 2);}

void factorial() {
    fac[0] = 1;
    for (int i = 1; i < N; i++) fac[i] = carp(fac[i - 1], i);
    invfac[N - 1] = inv(fac[N - 1]);
    for (int i = N - 2; i >= 0; i--)
        invfac[i] = carp(invfac[i + 1], i + 1);
}

int nCr(int m, int r) { return carp(fac[m], carp(invfac[r], invfac[m - r]));}

void solve() {
    cin >> n >> s;
    int z = 0, o = 0;
    for (int i = 0; i < n; i++) {
        int j = i;
        if (s[j] == '1') {
            while (j < n && s[j] == '1')
                j++;
            o += (j - i) / 2;
            i = j - 1;
        } else  z++;
    }
    cout << nCr(z + o, o) << nl;
}