AtCoder Grand Contest 036 简要题解
从这里开始
Problem A Triangle
考虑把三角形移到和坐标轴相交，即
然后能够用坐标比较简单地计算面积，简单构造一下就行了。
Code
#include <bits/stdc++.h>
using namespace std;
typedef bool bolean;
#define ll long long
const int V = 1e9;
ll S;
int main() {
scanf("%lld", &S);
if (S == 1000000000000000000ll) {
printf("%d 0 0 0 0 %d\n", V, V);
return 0;
}
int z = S / V;
int y = V;
int x = 1;
int h = S % V;
printf("0 %d %d 0 %d %d\n", y, x, x + z, h);
return 0;
}
Problem B Do Not Duplicate
你发现 $x$ 如果还会再出现，会把之间的数都删掉，所以 $x$ 向它的后继连一条边。然后模拟一下就行了。
Code
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;
#define ll long long
const int N = 2e5 + 5;
int n;
ll K;
int a[N], h[N];
int fa[N], c[N];
boolean vis[N];
int main() {
scanf("%d%lld", &n, &K);
for (int i = 0; i < n; i++) {
scanf("%d", a + i);
}
for (int i = n - 1; ~i; i--)
h[a[i]] = i + n;
for (int i = n - 1; ~i; i--) {
int suf = h[a[i]];
fa[i] = (suf + 1) % n;
c[i] = (suf + 1) / n;
h[a[i]] = i;
}
int p = 0, q;
vector<int> stk;
do {
vis[p] = true;
stk.push_back(p);
p = fa[p];
} while (!vis[p]);
int sumcost = 0;
do {
q = stk.back();
stk.pop_back();
sumcost += c[q];
} while (q ^ p);
stk.clear();
int x = 0;
while (K - c[x] >= 1 && (x ^ p))
K -= c[x], x = fa[x];
if (x == p)
K = (K - 1) % sumcost + 1;
while (K - c[x] >= 1)
K -= c[x], x = fa[x];
memset(vis, false, sizeof(vis));
for (int i = x; i < n + (K - 1) * n; i++) {
int z = a[i % n];
if (!vis[z]) {
stk.push_back(z);
vis[z] = true;
} else {
int y = 0;
do {
y = stk.back();
stk.pop_back();
vis[y] = false;
} while (y ^ z);
}
}
for (auto x : stk) {
printf("%d ", x);
}
return 0;
}
Problem C GP 2
考虑一些必要条件：
- 和为 $3M$
- 最大值不超过 $2M$
- 奇数个数不超过 $M$
可以对最大值和最大奇数或剩下的最大值进行讨论，然后用归纳法证明。
剩下是基础计数。
Code
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;
#define ll long long
void exgcd(int a, int b, int& x, int& y) {
if (!b) {
x = 1, y = 0;
} else {
exgcd(b, a % b, y, x);
y -= (a / b) * x;
}
}
int inv(int a, int n) {
int x, y;
exgcd(a, n, x, y);
return (x < 0) ? (x + n) : (x);
}
const int Mod = 998244353;
template <const int Mod = :: Mod>
class Z {
public:
int v;
Z() : v(0) { }
Z(int x) : v(x){ }
Z(ll x) : v(x % Mod) { }
friend Z operator + (const Z& a, const Z& b) {
int x;
return Z(((x = a.v + b.v) >= Mod) ? (x - Mod) : (x));
}
friend Z operator - (const Z& a, const Z& b) {
int x;
return Z(((x = a.v - b.v) < 0) ? (x + Mod) : (x));
}
friend Z operator * (const Z& a, const Z& b) {
return Z(a.v * 1ll * b.v);
}
friend Z operator ~(const Z& a) {
return inv(a.v, Mod);
}
friend Z operator - (const Z& a) {
return Z(0) - a;
}
Z& operator += (Z b) {
return *this = *this + b;
}
Z& operator -= (Z b) {
return *this = *this - b;
}
Z& operator *= (Z b) {
return *this = *this * b;
}
friend boolean operator == (const Z& a, const Z& b) {
return a.v == b.v;
}
};
Z<> qpow(Z<> a, int p) {
Z<> rt = Z<>(1), pa = a;
for ( ; p; p >>= 1, pa = pa * pa) {
if (p & 1) {
rt = rt * pa;
}
}
return rt;
}
typedef Z<> Zi;
const int N = 3e6 + 5;
int n, m;
Zi fac[N], _fac[N];
void prepare(int n) {
fac[0] = 1;
for (int i = 1; i <= n; i++)
fac[i] = fac[i - 1] * i;
_fac[n] = ~fac[n];
for (int i = n; i; i--)
_fac[i - 1] = _fac[i] * i;
}
Zi comb(int n, int m) {
return (n < m) ? (0) : (fac[n] * _fac[m] * _fac[n - m]);
}
int main() {
scanf("%d%d", &n, &m);
prepare(3 * m + n - 1);
Zi ans = comb(3 * m + n - 1, n - 1) - comb(m + n - 2, n - 1) * n;
for (int num = m + 1; num <= n && num <= 3 * m; num++) {
if ((3 * m - num) & 1)
continue;
int sum = (3 * m - num) >> 1;
Zi tmp = comb(sum + n - 1, n - 1);
// mx is odd
tmp -= num * comb(sum - m + n - 1, n - 1);
// mx is even
tmp -= (n - num) * comb(sum - m + n - 2, n - 1);
tmp *= comb(n, num);
ans -= tmp;
}
printf("%d\n", ans.v);
return 0;
}
Problem D Negative Cycle
设到第 $i$ 个点的最短路为 $d_i$。不难发现 $d_i \geqslant d_{i + 1}$，如果存在负环，那么会更新。
考虑最短路等于 $d$ 和最短路等于 $d + 1$ 的两段 $[a, b]$，以及 $(b, c]$。
那么 $(b, c]$ 不能有到 $a$ 之前的边，否则 $d$ 会被更新。$(b + 1, c]$ 之间不能有从前连向后面的边，否则 $d$ 也会被更新。
状态中记入 $a, b$，稍作优化就能做到 $O(n^3)$。
Code
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;
#define ll long long
const ll llf = (signed ll) (~0ull >> 2);
const int N = 505;
int n;
int A[N][N];
ll s[N][N];
ll dp[N][N];
int main() {
scanf("%d", &n);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i == j) {
continue;
}
scanf("%d", A[i] + j);
s[i][j] = A[i][j] + s[i - 1][j] + s[i][j - 1] - s[i - 1][j - 1];
}
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
dp[i][j] = llf;
dp[1][1] = 0;
for (int i = 2; i <= n; i++) {
dp[1][i] = dp[1][i - 1];
for (int j = 1; j < i; j++) {
dp[1][i] += A[j][i];
}
}
for (int i = 2; i <= n; i++) {
ll cost = 0;
for (int b = i - 1; b; b--) {
for (int j = b + 2; j <= i; j++)
cost += A[b + 1][j];
for (int a = b; a; a--) {
dp[b + 1][i] = min(dp[b + 1][i], dp[a][b] + cost + s[i][a - 1] - s[b][a - 1]);
}
}
}
ll ans = llf;
for (int i = 1; i <= n; i++)
ans = min(ans, dp[i][n]);
printf("%lld\n", ans);
return 0;
}
Problem E ABC String
先将连续的字符缩起来，设出现的次数满足 $a \leqslant b \leqslant c$。
现在我们希望删掉一些 $a$ 使得存在一种方案使得删掉一些 $b, c$ 使得 $a = b = c$。
不难发现，删掉一个 A 后至多使 $b$ 或 $c$ 减少 1。因此 A 仍然是出现次数最少的字符。
假设删除一些 A 之后，仍然满足 $b \leqslant c$。
如果 $b = c$，那么每次删掉 BC 或者 CB 就行了。
考虑 $b - c$ 所能到达的最小值，如果一段中含 B，那么可以贡献 -1，如果只有一个 C，那么会贡献 1。
因此，我们希望删掉一些 A，使得 $b - c$ 的最小值能小于等于 0。
不难发现，只用删掉若干个单独的 C 之前的 A，并且删掉一个 A 至多使单独的 C 减少 1 个。
Code
#include<bits/stdc++.h>
using namespace std;
typedef bool boolean;
const int N = 1e6 + 5;
int n;
char s[N], s1[N];
void shrink() {
int t = 0;
for (int i = 1; i <= n; i++) {
if (s[i] ^ s[i - 1]) {
s[++t] = s[i];
}
}
n = t;
}
int cnt[3];
char pg[3], qg[3];
void trans() {
for (int i = 1; i <= n; i++) {
cnt[s[i] - 'A']++;
}
vector<pair<int, int>> a;
a.emplace_back(cnt[0], 0);
a.emplace_back(cnt[1], 1);
a.emplace_back(cnt[2], 2);
sort(a.begin(), a.end());
#define make_trans(x, y) pg[x] = y, qg[y] = x;
for (int i = 0; i < 3; i++)
make_trans(a[i].second, i);
for (int i = 1; i <= n; i++) {
s[i] = pg[s[i] - 'A'] + 'A';
}
}
void make_valid() {
int x = 0, y = 0;
auto count = [&] (int l, int r) -> boolean {
if (l > r) return false;
for (int i = l; i <= r; i++) {
if (s[i] == 'B') {
x++;
return false;
}
}
if (l == 1 || r == n)
return false;
y++;
return true;
};
int ls = 0;
vector<int> pos;
for (int i = 1; i <= n; i++) {
if (s[i] == 'A') {
if (count(ls + 1, i - 1) && ls) {
pos.push_back(i);
}
ls = i;
}
}
count(ls + 1, n);
if (x < y) {
int t = 0;
pos.resize(y - x);
reverse(pos.begin(), pos.end());
for (int i = 1; i <= n; i++) {
if (!pos.empty() && i == pos.back()) {
pos.pop_back();
} else {
s[++t] = s[i];
}
}
n = t;
shrink();
s[n + 1] = 0;
}
memset(cnt, 0, sizeof(cnt));
for (int i = 1; i <= n; i++)
cnt[s[i] - 'A']++;
int d = cnt[2] - cnt[1];
int t = 0;
char target = 'C';
(d < 0) && (d = -d, target = 'B');
ls = 0;
auto remove = [&] (int l, int r) {
if (l > r)
return;
if ((l == 1 || r == n) || (l ^ r)) {
if (s[l] == target && d) l++, d--;
if (l < r && s[r] == target && d) r--, d--;
}
for (int i = l; i <= r; i++)
s1[++t] = s[i];
};
for (int i = 1; i <= n; i++) {
if (s[i] == 'A') {
remove(ls + 1, i - 1);
s1[++t] = 'A';
ls = i;
}
}
remove(ls + 1, n);
n = t;
for (int i = 1; i <= n; i++) {
s[i] = s1[i];
}
s[n + 1] = 0;
}
void make_equal() {
memset(cnt, 0, sizeof(cnt));
for (int i = 1; i <= n; i++)
cnt[s[i] - 'A']++;
assert(cnt[1] == cnt[2]);
int d = cnt[1] - cnt[0];
int t = 0;
vector<pair<int, int>> pos;
for (int i = 1; i <= n; i++) {
if (i < n && d && s[i] != 'A' && s[i + 1] != 'A') {
if (s1[t] == 'A' && s[i + 2] == 'A') {
pos.emplace_back(t + 1, t + 2);
s1[++t] = s[i];
} else {
i++, d--;
}
} else {
s1[++t] = s[i];
}
}
n = t;
for (int i = 1; i <= n; i++)
s[i] = s1[i];
vector<int> delpos;
for (int i = 0; i < d; i++) {
delpos.push_back(pos[2 * i].first);
delpos.push_back(pos[2 * i + 1].second);
}
t = 0;
reverse(delpos.begin(), delpos.end());
for (int i = 1; i <= n; i++) {
if (!delpos.empty() && delpos.back() == i) {
delpos.pop_back();
} else {
s[++t] = s[i];
}
}
t = n;
assert(delpos.empty());
}
int main() {
scanf("%s", s + 1);
n = strlen(s + 1);
shrink();
trans();
make_valid();
make_equal();
for (int i = 1; i <= n; i++)
s[i] = qg[s[i] - 'A'] + 'A';
s[n + 1] = 0;
puts(s + 1);
return 0;
}
Problem F Square Constraints
考虑每个位置有一个上界 $R_i$ 和一个下界 $L_i$。对于后 $N$ 个位置下界为 0。
考虑只有上界，我们把上界从小到大排序，那么答案等于 $\prod_{i = 0}^{2N - 1} (R_i - i)$。
考虑对一些下界进行容斥。不难注意到下界非 0 的位置，它的上界是排在后 $N$ 个位置的上界之后。
所以只用枚举一下有多少个数被硬点不合法就可以 dp 了。
Code
#include <bits/stdc++.h>
using namespace std;
typedef bool boolean;
#define pii pair<int, int>
#define ll long long
const int N = 505;
int n, n2, Mod;
pii a[N];
int cnt[N];
int dp[2][N];
int main() {
scanf("%d%d", &n, &Mod);
n2 = n << 1;
int p1 = 0, p2 = 0, N4 = 4 * n * n, N2 = n * n;
for (int i = n2 - 1; ~i; i--) {
while (p1 < n2 && p1 * p1 <= N4 - i * i)
p1++;
while (p2 * p2 < N2 - i * i)
p2++;
if (p2) {
a[i] = pii(p2, p1);
} else {
a[i] = pii(p1, 0);
}
}
sort(a, a + n2);
cnt[0] = 0;
for (int i = 0; i < n2; i++)
cnt[i + 1] = cnt[i] + !a[i].second;
ll ans = 0;
for (int i = 0; i <= n; i++) {
int cur = 0;
memset(dp[0], 0, sizeof(dp[0]));
dp[0][0] = 1;
for (int j = 0; j < n2; j++) {
memset(dp[cur ^= 1], 0, sizeof(dp[0]));
for (int k = 0; k <= i; k++) {
if (!a[j].second) {
dp[cur][k] = (dp[cur][k] + 1ll * (a[j].first - k - cnt[j]) * dp[cur ^ 1][k]) % Mod;
} else {
dp[cur][k + 1] = (dp[cur][k + 1] - 1ll * (a[j].first - k - cnt[j]) * dp[cur ^ 1][k]) % Mod;
dp[cur][k] = (dp[cur][k] + 1ll * (a[j].second - (n + i + j - cnt[j] - k)) * dp[cur ^ 1][k]) % Mod;
}
}
}
ans += dp[cur][i];
}
ans %= Mod;
(ans < 0) && (ans += Mod);
printf("%d\n", (int) ans);
return 0;
}
