AtCoder Beginner Contest 323

1|0A - Weak Beats

#include <bits/stdc++.h>

using namespace std;

using pii = pair<int, int>;
using vi = vector<int>;


int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    string s;
    cin >> s;
    for( int i = 1 ; i < 16 ; i += 2 )
        if( s[i] == '1' ){
            cout << "No\n";
            return 0;
        }
    cout << "Yes\n";
    return 0;
}

2|0B - Round-Robin Tournament

#include <bits/stdc++.h>

using namespace std;

using pii = pair<int, int>;
using vi = vector<int>;


int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    string s;
    int n;
    cin >> n;
    vector<pii> a;
    for (int i = 1, cnt; i <= n; i++) {
        cin >> s, cnt = 0;
        for (auto j: s)
            cnt += (j == 'o');
        a.emplace_back(cnt, i);
    }
    sort( a.begin(), a.end() , []( pii x , pii y ){
       if( x.first != y.first ) return x.first > y.first;
       return x.second < y.second;
    });
    for( auto [ cnt , id ] : a )
        cout << id << " ";

    return 0;
}

3|0C - World Tour Finals

#include <bits/stdc++.h>

using namespace std;

using pii = pair<int, int>;
using vi = vector<int>;


int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vi a(m), cnt(n);
    for (auto &i: a) cin >> i;
    for (int i = 0; i < n; i++)
        cnt[i] = i + 1;
    vector<string> s(n);
    for (auto &i: s)
        cin >> i;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++)
            cnt[i] += (s[i][j] == 'o') * a[j];
    int top = *max_element(cnt.begin(), cnt.end());
    for (int i = 0, res; i < n; i++) {
        priority_queue<int> q;
        for (int j = 0; j < m; j++)
            if (s[i][j] == 'x') q.push(a[j]);
        res = 0;
        while (!q.empty() and cnt[i] < top)
            cnt[i] += q.top(), q.pop(), res++;
        cout << res << "\n";
    }
    return 0;
}

4|0D - Merge Slimes

#include <bits/stdc++.h>

using namespace std;

#define int long long
using pii = pair<int, int>;
using vi = vector<int>;


int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;
    map<int, int> a;
    for (int x, y; n; n--)
        cin >> x >> y, a[x] += y;
    for (auto &[c, s]: a) {
        a[c * 2] += s / 2, s %= 2;
    }
    int res = 0;
    for( auto [c,s] : a )
        res += s;
    cout << res << "\n";
    return 0;
}

5|0E - Playlist

令 $f[i]$ 表示第 $i$ 秒开始播放音乐的概率，所以 $\frac 1 n \sum_{i=x-t_0+1}^{x}f[i]$ 就是答案。

我们求可以枚举上一首放的歌，然后从转移过来即可。记得要乘上播放这首歌的概率 $\frac 1 n$

#include <bits/stdc++.h>

using namespace std;

#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;
const int mod = 998244353;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vi t(n);
    for (auto &i: t) cin >> i;
    vi f(m + 1);
    f[0] = 1;
    int inv = power(n, mod - 2);
    for (int i = 1; i <= m; i++) {
        for (auto j: t) {
            if (i - j < 0) continue;
            f[i] = (f[i] + f[i - j]) % mod;
        }
        f[i] = f[i] * inv % mod;
    }
    int ans = 0;
    for (int i = max(0ll, m - t[0] + 1); i <= m ; i++)
        ans = (ans + f[i]) % mod;
    ans = ans * inv % mod;
    cout << ans << "\n";
    return 0;
}

6|0F - Push and Carry

首先我们一定是是移动到某一个点上，然后把货物一直推到终点。

如果货物到终点是一条斜线，则过程中需要变换一次方向，移动次数加 2

如果人货物终点三点共线，且人位于中间，则需要绕道另一侧，移动次数加 2

然后就是特判出移动到货物周围的哪一个点。

#include <bits/stdc++.h>

using namespace std;

#define int long long
using i32 = int32_t;
using vi = vector<int>;
using pii = pair<int, int>;

const int inf = 1e9, INF = 1e18;
const int mod = 998244353;

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int xa, xb, xc, ya, yb, yc;
    cin >> xa >> ya >> xb >> yb >> xc >> yc;
    xa -= xb, ya -= yb, xc -= xb, yc -= yb, xb = yb = 0;
    int res = abs(xc) + abs(yc);

    if (xc < 0 and yc < 0) xc = -xc, yc = -yc, xa = -xa, ya = -ya;
    if (xc > 0 and yc > 0) {
        cout << res + min(abs(xa + 1) + abs(ya), abs(xa) + abs(ya + 1)) + 2;
        return 0;
    }

    if (xc > 0 and yc < 0) xc = -xc, yc = -yc, xa = -xa, ya = -ya;
    if (xc < 0 and yc > 0) {
        cout << res + min(abs(xa - 1) + abs(ya), abs(xa) + abs(ya + 1)) + 2;
        return 0;
    }
    if (xc == 0) {
        if (yc < 0) xc = -xc, yc = -yc, xa = -xa, ya = -ya;
        if (ya > 0 and xa == 0) res += 2;
        cout << res + abs(xa) + abs(ya + 1);
    } else {
        if (xc < 0) xc = -xc, yc = -yc, xa = -xa, ya = -ya;
        if (xa > 0 and ya == 0) res += 2;
        cout << res + abs(xa + 1) + abs(ya);
    }
    return 0;
}