Codeforces Round #604 (Div. 1)
Codeforces Round #604 (Div. 1)
A
先把原序列去重弄成二元组 \((x,c_x)\)，即解了 \(x\) 题的有 \(c_x\) 人。这样的处理方便解决“严格大于”的限制，只要每种奖牌取一些二元组即可。
然后找到人数一半的位置把后面的仍掉。接下来枚举金牌的人数，然后从后往前截取尽可能短的一段当作铜牌（需满足人数多于金牌），看看中间的一段人数是否多于金牌人数。如果多了就说明满足条件，更新一下答案
#include <bits/stdc++.h>
using namespace std;
void read (int &x) {
char ch = getchar(); x = 0; while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 4e5 + 5;
int n, a[N], c[N];
struct qwq { int x, c; } b[N];
signed main() {
int T; read (T);
while (T--) {
read (n);
for (int i = 1; i <= n; ++i) read (a[i]);
int m = 0;
for (int i = 1; i <= n; ++i)
if (a[i] == a[i - 1]) ++b[m].c;
else b[++m].x = a[i], b[m].c = 1;
int p, x = 0, y = 0, z = 0, s = 0, o = 0;
for (p = 1; p <= m; ++p) {
s += b[p].c; if (s * 2 > n) break;
} --p;
for (int i = 1; i <= p; ++i) c[i] = c[i - 1] + b[i].c;
for (int i = 1, j = p; i < j; ++i) {
int tx = c[i];
while (c[p] - c[j - 1] <= tx && j > i) --j;
if (i == j) break;
int ty = c[j - 1] - c[i], tz = c[p] - c[j - 1];
if (ty > tx && tx + ty + tz > o)
o = tx + ty + tz, x = tx, y = ty, z = tz;
}
printf ("%d %d %d\n", x, y, z);
}
}
B
抓住 \(0\) 和 \(3\) 只能分别和一种数相邻的性质。“\(01\)” 可以连续相间分布，中间夹有的其他数字移到两边不影响结果。于是方案可以描述为 \((1),0,1,0,1,...0,1|2,1,2,1,...2,1,|2,3,2,3,...2,3,(2)\)
设 \(one=b-a\)，\(two=c-d\)，由于两头可以放或不放 \(1,2\)，所以优解的条件是 \(|one-two|\leq 1 \&one,two\ge0\)。但还有一个特殊情况，只有 \(0,1\) 或只有 \(2,3\)，且 \(one=-1\) 或 \(two=-1\)，这时在两头都放 \(0\) 或 \(3\) 也合法。然后按上述序列构造即可
#include <bits/stdc++.h>
using namespace std;
void read (int &x) {
char ch = getchar(); x = 0; while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 1e5 + 5;
#define pc(x) putchar (x), putchar (' ')
signed main() {
int a, b, c, d;
read (a), read (b), read (c), read (d);
b -= a, c -= d;
if (b >= 0 && c >= 0 && abs (b - c) <= 1) {
puts ("YES");
if (b > c) pc ('1');
for (int i = 1; i <= a; ++i) pc ('0'), pc ('1');
for (int i = 1; i <= min (b, c); ++i) pc ('2'), pc ('1');
for (int i = 1; i <= d; ++i) pc ('2'), pc ('3');
if (b < c) pc ('2');
} else if (a == 0 && b == 0 && c == -1) {
puts ("YES");
pc ('3'); for (int i = 1; i < d; ++i) pc ('2'), pc ('3');
} else if (c == 0 && d == 0 && b == -1) {
puts ("YES");
for (int i = 1; i < a; ++i) pc ('0'), pc ('1'); pc ('0');
}
else puts ("NO");
}
C
按照检查点分成几段，期望天数显然为每段之和。那么每一段的答案怎么算呢？
设 \(E(x)\) 为从 \(x\) 开始到结束的期望步数，有 \(E(x)=P_xE(x+1)+(1-P_x)E(c)+1\)，\(c\) 为 \(x\) 之前的最接近的检查点，\(c'\) 为后面的最近的检查点。一个长度为 \(k\) 的段可以写出 \(k\) 个式子（方程），对这个方程组进行消元，得到 \(E(c')-E(c)=\frac{1+\sum_{i=1}^{k}\prod_{j=1}^{i}P_j}{\prod_{i=1}^{k} P_i}\)，即这一段的期望值。维护一些前缀和即可 \(O(1)\) 计算。段与段之间会多算一次，还要减去段的数量
而每次修改只会影响 \(O(1)\) 的区间，\(set\) 维护一下就好了。
#include <bits/stdc++.h>
using namespace std;
#define int long long
void read (int &x) {
char ch = getchar(); x = 0; while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 2e5 + 5, mod = 998244353;
int qpow (int x, int y) {
int t = 1; for (; y; y >>= 1, x = x * x % mod) if (y & 1) t = t * x % mod; return t;
}
int n, q, res, a[N], s[N]; set<int> t; set<int>::iterator it;
#define inv(x) qpow (x, mod - 2)
int f (int l, int r) {
return ((s[r] - s[l - 1] + mod) * inv (a[r]) + a[l - 1] * inv (a[r])) % mod;
}
#define up(x, y) ((x += y) >= mod && (x -= mod))
signed main() {
read (n), read (q); a[0] = 1;
for (int i = 1; i <= n; ++i)
read (a[i]), a[i] = a[i] * qpow (100 , mod - 2) % mod;
for (int i = 2; i <= n; ++i) a[i] = a[i] * a[i - 1] % mod;
for (int i = 1; i <= n; ++i) s[i] = (s[i - 1] + a[i]) % mod;
res = f (1, n), t.insert (1), t.insert (n + 1); int num = 1;
for (int i = 1, x, l, r; i <= q; ++i) {
read (x); it = t.find (x);
if (it == t.end()) {
t.insert (x); it = t.find (x); ++num;
--it; l = *it; ++it, ++it, r = *it;
up (res, mod - f (l, r - 1));
up (res, f (l, x - 1)), up (res, f (x, r - 1));
} else {
--it, l = *it; ++it, ++it, r = *it;
up (res, mod - f (l, x - 1));
up (res, mod - f (x, r - 1));
up (res, f (l, r - 1));
--it, t.erase (it); --num;
}
printf ("%lld\n", (res + mod - num) % mod);
}
}
D
考虑每个 ‘(’ 有多少次被算入答案中。如果 \(x\) 被算入，需满足 \(x\) 左边的 ‘(' 的数量小于右边 ')' 的数量，由此列出算式：\(a\) 为左侧 ‘(' 的数量，\(b\) 为左侧 ‘?' 的数量；\(c\) 为右侧 ‘)' 的数量，\(d\) 为右侧 ‘?' 的数量，可以枚举左侧 ‘?’ 变为 ‘(‘ 的数量和右侧 '?' 变成 ')' 的数量，有 \(number=\sum_{a+i<c+j}C(b,i)·C(d,j)\)。尽量把 \(i,j\) 在限制不等式中弄成同号方便合并，于是用 \(d-j\) 换掉 \(j\)。变成了这样 \(number=\sum_{i+j<c+d-a}C(b,i)·C(d,j)\)。在 \(b\) 中选 \(i\) 个，\(d\) 中选 \(j\) 个，相当于在 \(b+d\) 中选 \(i+j\) 个。这样就可以合并了，用 \(k\) 换掉 \(i+j\)，得\(res=\sum_{k<c+d-a}C(b+d,k)\)
\(b+d\) 为左右 '?' 的总数，最多只有两种取值，因此可以预处理这两种组合数的前缀和，然后加一加
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e6 + 5, mod = 998244353;
int n, in[N], p[N], s[2][N]; char a[N];
int qpow (int x, int y) {
int t = 1; for (; y; y >>= 1, x = x * x % mod) if (y & 1) t = t * x % mod; return t;
}
int C (int x, int y) {
return x < y ? 0 : p[x] * in[y] % mod * in[x - y] % mod;
}
void prework () {
p[0] = 1; for (int i = 1; i <= n; ++i) p[i] = p[i - 1] * i % mod;
in[n] = qpow (p[n], mod - 2);
for (int i = n; i >= 1; --i) in[i - 1] = in[i] * i % mod;
}
int cl[N], cr[N], sl[N], sr[N];
signed main() {
scanf ("%s", a + 1); n = strlen (a + 1); prework ();
for (int i = 1; i <= n; ++i)
cl[i] = cl[i - 1] + (a[i] == '('), sl[i] = sl[i - 1] + (a[i] == '?');
for (int i = n; i >= 1; --i)
cr[i] = cr[i + 1] + (a[i] == ')'), sr[i] = sr[i + 1] + (a[i] == '?');
s[0][0] = s[1][0] = 1; int res = 0;
for (int i = 1; i <= n; ++i) {
s[1][i] = (s[1][i - 1] + C (sl[n] - 1, i)) % mod;
s[0][i] = (s[0][i - 1] + C (sl[n], i)) % mod;
}
for (int i = 1; i <= n; ++i) if (cl[i - 1] < cr[i + 1] + sr[i + 1] && a[i] != ')')
(res += s[a[i] == '?'][cr[i + 1] + sr[i + 1] - cl[i - 1] - 1]) %= mod;
printf ("%lld\n", res);
}
E
先反其道而行之，如果 \((A,B,C)\) 不符合条件，一定有某一个队伍赢了另外两个队。进一步，如果 \(x\) 队赢了 \(c_x\) 局，不满足条件的三元组数量 \(number=\sum\frac{c_x(c_x-1)}{2}\)。目的就是让这个东西尽量小
现在有一些的比赛结果已经固定，相当于每个 \(c_x\) 有一个限制：\(c_x\in [l_x,r_x]\) 且 \(\sum c_x=\frac{n(n-1)}{2}\)。如果没有这个限制当然可以根据不等式直接取值，但加上限制后似乎没法直接贪心。（不然为什么数据范围这么小呢）
改用费用流。每场比赛一个节点，每个队伍一个节点。连这几种边：
1、\(S\) 到所有比赛节点连一条容量为 \(1\)，花费为 \(0\) 的边，代表这场比赛赢家只有一个。
2、每场比赛向对应的两支队伍连边，容量 \(1\)，花费 \(0\)
3、每支队伍向 \(T\) 连 \(n-p_x-1\) 条边，\(p_x\) 为固定的已经赢的场数。每条边的容量为 \(1\)，花费为每次多赢一场后 \(\frac{c_x(c_x-1)}{2}\) 的增量
在这张图上求最小费用最大流就是答案
#include <bits/stdc++.h>
using namespace std;
void read (int &x) {
char ch = getchar(); x = 0; while (!isdigit(ch)) ch = getchar();
while (isdigit(ch)) x = x * 10 + ch - 48, ch = getchar();
} const int N = 52, P = N * N, M = N * 600;
int n, m, S, T, tot, k[N][N], c[N], idm[N][N], id[N];
int cnt = 1, h[M], nxt[M], to[M], w[M], co[M];
void add (int u, int v, int ww, int cc) {
to[++cnt] = v, w[cnt] = ww, co[cnt] = cc, nxt[cnt] = h[u], h[u] = cnt;
to[++cnt] = u, w[cnt] = 0, co[cnt] = -cc, nxt[cnt] = h[v], h[v] = cnt;
}
queue<int> q; int d[P], f[P], in[P], pre[P];
int spfa () {
memset (d, 0x3f, sizeof (d));
memset (in, 0, sizeof (in));
q.push (S), d[S] = 0, f[S] = 2e9;
while (!q.empty()) {
int u = q.front (); q.pop (), in[u] = 0;
for (int i = h[u], v; i; i = nxt[i])
if (w[i] && d[v = to[i]] > d[u] + co[i]) {
d[v] = d[u] + co[i];
f[v] = min (f[u], w[i]); pre[v] = i;
if (!in[v]) in[v] = 1, q.push (v);
}
}
return (d[T] != d[T + 1]);
}
void update () {
for (int x = T, i; x != S; x = to[i ^ 1]) {
i = pre[x]; w[i] -= f[T], w[i ^ 1] += f[T];
}
}
signed main() {
read (n), read (m);
for (int i = 1, a, b; i <= m; ++i)
read (a), read (b), k[a][b] = 1, ++c[a];
for (int i = 1; i <= n; ++i) id[i] = ++tot;
for (int i = 1; i <= n; ++i)
for (int j = i + 1; j <= n; ++j)
if (!k[i][j] && !k[j][i]) {
idm[i][j] = ++tot; add (0, tot, 1, 0);
add (tot, id[i], 1, 0); add (tot, id[j], 1, 0);
}
S = 0, T = ++tot;
for (int i = 1; i <= n; ++i)
for (int j = c[i]; j < n; ++j) add (id[i], T, 1, j);
while (spfa ()) update ();
for (int i = 1; i <= n; ++i)
for (int j = i + 1; j <= n; ++j)
if (idm[i][j]) {
int x = idm[i][j], tag = 0, y = 0;
for (int k = h[x]; k; k = nxt[k])
if (to[k] == 0 && w[k] == 1) tag = 1;
if (!tag) continue;
for (int k = h[x]; k; k = nxt[k])
if (to[k] && w[k] == 0) y = to[k];
y == i ? k[i][j] = 1 : k[j][i] = 1;
}
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) putchar (k[i][j] + '0'); puts ("");
}
}
