2022 International Collegiate Programming Contest, Jinan Site
目录
VP概况
没有罚时的情况下拿下签到，E因为打的表太小，猜错了结论，后面让队友读完大模拟来写，也很顺利，看榜发现后面的题不可做，补题的时候发现也不是1小时可以写出来的，下次比赛按照打满的思路来
M - Best Carry Player
推导得知运算顺序不影响答案，直接模拟
int n;
void solve()
{
cin>>n;
vector<int> a, b, res;
int x; cin>>x;
int ans = 0;
while(x)
{
a.push_back(x % 10);
x /= 10;
}
for(int i = 2; i <= n; i++)
{
cin>>x;
b.clear();
res.clear();
while(x)
{
b.push_back(x % 10);
x /= 10;
}
int len1 = a.size(), len2 = b.size();
int add = 0;
for(int j = 0; j < max(len1, len2); j++)
{
int t1 = 0, t2 = 0;
if(j < len1)
t1 = a[j];
if(j < len2)
t2 = b[j];
int s = t1 + t2 + add;
// cout<<s<<'\n';
res.push_back(s % 10);
// cout<<i<<" "<<j<<" "<<add<<'\n';
add = s / 10;
if(add == 1)
ans++;
}
if(add)
res.push_back(add);
swap(a, res);
// a = res;
}
cout<<ans<<'\n';
}
K - Stack Sort
看有多少个相邻段在数值上和数组下标上是连续的
const int N = 5e5 + 10;
int n, a[N], pos[N];
void solve()
{
cin>>n;
int res = 0;
for(int i = 1; i <= n; i++)
{
cin>>a[i];
pos[a[i]] = i;
}
for(int i = n; i >= 1; i--)
{
int j = i;
while(j >= 2 && pos[j - 1] >= pos[j])
j--;
res++;
i = j;
}
cout<<res<<'\n';
return;
}
A - Tower
如果所有数字都要变成 \(x\) ，那么就是一个中位数问题了，其区别在于除 \(2\) 的操作，除 \(2\) 代表可以有更多的中位数选择，那么可能得中位数数量有 \(O(N \log V)\) 种，然后去判断每个数到达 \(x\) 的最小操作次数，总共时间复杂度 \(N^2 \log V \times (\log V + \log N)\) ，这里包括排序的复杂度
int n, m, a[510];
vector<int> b;
ll check(int x)
{
ll ans = 0;
vector<int> c;
// cout<<"X : ";
for(int i = 1; i <= n; i++)
{
if(a[i] <= x)
{
c.push_back(x - a[i]);
// cout<<x - a[i]<<" ";
continue;
}
int y = a[i];
int cost = a[i] - x;
int t = 0;
while(y)
{
cost = min(abs(y - x) + t, cost);
y /= 2;
t++;
}
c.push_back(cost);
}
sort(c.begin(), c.end());
reverse(c.begin(), c.end());
for(int i = m; i < n; i++)
ans += c[i];
return ans;
}
void solve()
{
b.clear();
cin>>n>>m;
for(int i = 1; i <= n; i++)
{
cin>>a[i];
int x = a[i];
while(x)
{
b.push_back(x);
x /= 2;
}
}
b.push_back(0);
sort(b.begin(), b.end());
b.erase(unique(b.begin(), b.end()), b.end());
ll res = 1e18;
for(auto it : b)
res = min(check(it), res);
cout<<res<<'\n';
}
E - Identical Parity
我是打表找规律做的，队内的两位数学手没做出来，我先打表打出来了
规律如下，表中 \(1\) 代表 YES在 \(k\) 为奇数的时候，YES 情况成等差数列分布
\(k = 1, 10\)
\(k = 3, 111010\)
\(k = 5, 11111011100010\)
\(\dots\)
\(1\) 和 \(0\)的数量分部刚好等于 \(k + 1\) ，就很好的去做这个规律了
k: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
n : 1 1
n : 2 0 1
n : 3 0 1 1
n : 4 0 1 1 1
n : 5 0 1 1 1 1
n : 6 0 1 0 1 1 1
n : 7 0 1 1 1 1 1 1
n : 8 0 1 0 1 1 1 1 1
n : 9 0 1 0 1 1 1 1 1 1
n : 10 0 1 0 1 0 1 1 1 1 1
n : 11 0 1 0 1 1 1 1 1 1 1 1
n : 12 0 1 0 1 1 1 1 1 1 1 1 1
n : 13 0 1 0 1 1 1 1 1 1 1 1 1 1
n : 14 0 1 0 1 0 1 0 1 1 1 1 1 1 1
n : 15 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1
n : 16 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1
n : 17 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 18 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1
n : 19 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 20 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 21 0 1 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 22 0 1 0 1 0 1 0 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1
n : 23 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 24 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
n : 25 0 1 0 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
打表代码（丢失了）
AC代码：
typedef long long ll;
void solve()
{
ll n, k;
cin>>n>>k;
if(n == 1)
{
cout<<"YES\n";
return;
}
if(k == 1)
{
cout<<"NO\n";
return;
}
if(k % 2 == 0)
{
cout<<"YES\n";
return;
}
ll m = (n + 1) / 2;
ll x = k;
ll t = (n - (k - 1)) / (x + 1), r = (n - (k - 1)) % (x + 1);
if(r != 0) t++;
if(r == 0) r = x + 1;
ll d = 1ll + (t - 1) * 2;
// cout<<"t: "<<t<<" m: "<<m<<" r: "<<r<<" d: "<<d<<"\n";
if(t <= m && r <= x + 1 - d)
cout<<"YES\n";
else
cout<<"NO\n";
}
D - Frozen Scoreboard
大模拟
对于要求的总时间先减去已经确定的时间，即是封榜前AC的题的罚时，再枚举 \(S\) 其二进制下第 \(i\) 位为 \(1\) 代表通过第 \(i - 1\) 题，对于状态 \(S\) 封榜后通过的题，总时间减去\(240\) 的基础时间，再用罚时消耗总时间，剩下的时间再用封榜后的60min消耗
并不难，细节多而已，代码的注释更清楚
array<int, 3> a[15];
array<int, 4> res[15];
int n, m;
void solve()
{
int ac, time; cin>>ac>>time;
for(int i = 0; i < m; i++)
{
char opt; cin>>opt;
if(opt == '.')
{
res[i] = {0, 0, 0, 0};
a[i] = {0, 0, 0};
}
else if(opt == '+')
{
int x, y; cin>>x>>opt>>y;
time -= ((x - 1) * 20 + y);
a[i] = {1, x, y};
res[i] = {1, x, y, 0};
}
else if(opt == '?')
{
int x, y; cin>>x>>y;
a[i] = {2, x, y};
res[i] = {2, 0, 0, 0};
}
else if(opt == '-')
{
int x; cin>>x;
a[i] = {3, 0, x};
res[i] = {3, 0, x, 0};
}
}
// cout<<time<<'\n';
for(int S = 0; S < (1 << m); S++)
{
bool ok = true;
int cnt = 0;
for(int i = 0; i < m; i++)
{
if((a[i][0] == 0 || a[i][0] == 3) && ((S >> i) & 1) == 0)
ok = false;
if(a[i][0] == 1 && ((S >> i) & 1) == 0)
ok = false;
}
for(int i = 0; i < m; i++)
{
if(a[i][0] == 1) cnt++;
else if(a[i][0] == 2 && ((S >> i) & 1) == 1) cnt++;
}
if(!ok || ac != cnt) continue;
int T = time;
// 过题时间 罚时次数
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
{
T -= 240;
T -= (a[i][2] - a[i][1]) * 20;
}
if(T < 0) continue;
// T 基础罚时
// 优先消耗封榜的罚时 20 倍数
// cout<<T<<'\n';
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
{
int c = min(a[i][1] - 1, T / 20);
T -= c * 20;
}
// 消耗罚时时间
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
T -= min(59, T);
if(T == 0) //输出答案
{
T = time;
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
{
T -= 240;
T -= (a[i][2] - a[i][1]) * 20;
}
// T 基础罚时
// 优先消耗封榜的罚时 20 倍数
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
{
int c = min(a[i][1] - 1, T / 20);
T -= c * 20;
res[i] = {1, a[i][2] - a[i][1] + c + 1, 240, 0};
// 封榜前次数 封榜后的第几次
}
// 消耗罚时时间
for(int i = 0; i < m; i++)
if(((S >> i) & 1) && a[i][0] == 2)
{
res[i][0] = 1;
res[i][2] += min(59, T);
T -= min(59, T);
}
cout<<"Yes\n";
for(int i = 0; i < m; i++)
{
if(a[i][0] == 0)
cout<<".\n";
else if(a[i][0] == 1)
cout<<"+ "<<res[i][1]<<"/"<<res[i][2]<<'\n';
else if(a[i][0] == 2 && ((S >> i) & 1) == 1)
cout<<"+ "<<res[i][1]<<"/"<<res[i][2]<<'\n';
else if(a[i][0] == 2 && ((S >> i) & 1) == 0)
cout<<"- "<<a[i][2]<<"\n";
else
cout<<"- "<<a[i][2]<<"\n";
}
return;
}
}
cout<<"No\n";
}
G - Quick Sort
因为递归次数只有 \(\log n\) 次，则交换次数最多 \(n \log n\) 次，所以可以直接做
每个数字只会出现 \(1\) 次，用数据结构维护区间最大值，区间最小值即可
细节要注意的地方：par函数内，找到的下标不在范围内，那么我们就要找更大或更小的，为什么呢？因为 \(pivot\) swap了，导致 \(\geq\) 或 \(\leq\) \(pivot\)不在范围内
const int N = 5e5 + 10;
int n, a[N], res;
struct segtree
{
int w1, w2;
}seg[N * 4];
void update(int id)
{
seg[id].w1 = max(seg[id * 2].w1, seg[id * 2 + 1].w1);
seg[id].w2 = min(seg[id * 2].w2, seg[id * 2 + 1].w2);
}
void build(int id, int l, int r)
{
seg[id] = {0, 0};
if(l == r)
{
seg[id] = {a[l], a[l]};
return;
}
int mid = (l + r) >> 1;
build(id * 2, l, mid);
build(id * 2 + 1, mid + 1, r);
update(id);
}
void change(int id, int l, int r, int pos)
{
if(l == r)
{
seg[id] = {a[l], a[l]};
return;
}
int mid = (l + r) >> 1;
if(pos <= mid)
change(id * 2, l, mid, pos);
else
change(id * 2 + 1, mid + 1, r, pos);
update(id);
}
int query1(int id, int l, int r, int val)
{
if(l == r)
return l;
int mid = (l + r) >> 1;
if(seg[id * 2].w1 >= val)
return query1(id * 2, l, mid, val);
else
return query1(id * 2 + 1, mid + 1, r, val);
}
int query2(int id, int l, int r, int val)
{
if(l == r)
return l;
int mid = (l + r) >> 1;
if(seg[id * 2 + 1].w2 <= val)
return query2(id * 2 + 1, mid + 1, r, val);
else
return query2(id * 2, l, mid, val);
}
int par(int l, int r)
{
int tl = l - 1, tr = r + 1;
int val = a[(l + r) / 2];
while(1)
{
int p1 = query1(1, 1, n, val);
if(p1 <= tl) p1 = query1(1, 1, n, val + 1);
int p2 = query2(1, 1, n, val);
if(p2 >= tr) p2 = query2(1, 1, n, val - 1);
if(p1 >= p2)
return p2;
swap(a[p1], a[p2]);
change(1, 1, n, p1); change(1, 1, n, p2);
res++;
tl = p1, tr = p2;
}
}
void quicksort(int l, int r)
{
if(l >= r || l < 0 || r < 0) return;
int p = par(l, r);
quicksort(l, p);
quicksort(p + 1, r);
}
void solve()
{
cin>>n;
for(int i = 1; i <= n; i++)
cin>>a[i];
res = 0;
build(1, 1, n);
quicksort(1, n);
cout<<res<<'\n';
return;
}
C - DFS Order 2
回退背包，第一次见耶
2022 ICPC 济南站 C (回退背包) - 严格鸽的文章 - 知乎
typedef long long ll;
const int mod = 998244353;
const int N = 500 + 10;
ll qmi(ll a, ll b, ll mod)
{
ll ans = 1 % mod;
while(b)
{
if(b & 1) ans = ans * a % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
vector<int> e[N];
int n;
ll son[N], sz[N], fac[N];
ll f[N][N], g[N][N], h[N];
ll dfs1(int u, int from)
{
son[u] = 0;
sz[u] = 1;
ll res = 1;
for(auto v : e[u])
{
if(v == from) continue;
res = (res * dfs1(v, u)) % mod;
sz[u] += sz[v];
son[u]++;
}
res = (res * fac[son[u]]) % mod;
return res;
}
void dfs(int u, int from)
{
memset(g, 0, sizeof g);
g[0][0] = 1;
for(auto v : e[u])
{
if(v == from) continue;
for(int i = son[u]; i >= 1; i--)
for(int j = sz[u]; j >= sz[v]; j--)
g[i][j] = (g[i][j] + g[i - 1][j - sz[v]]) % mod;
}
for(auto v : e[u])
{
if(v == from) continue;
for(int i = 1; i <= son[u]; i++)
for(int j = sz[v]; j <= sz[u]; j++)
g[i][j] = (g[i][j] - g[i - 1][j - sz[v]] + mod) % mod;
memset(h, 0, sizeof h);
for(int i = 0; i <= son[u] - 1; i++)
for(int k = 0; k <= sz[u]; k++)
h[k + 1] = (h[k + 1] + (fac[i] * fac[son[u] - 1 - i] % mod) * g[i][k]) % mod;
for(int i = 1; i <= n; i++)
for(int k = 1; k <= n; k++)
if(i + k <= n)
f[v][i + k] = (f[v][i + k] + f[u][i] * h[k]) % mod;
for(int i = son[u]; i >= 1; i--)
for(int j = sz[u]; j >= sz[v]; j--)
g[i][j] = (g[i][j] + g[i - 1][j - sz[v]]) % mod;
}
for(auto v : e[u])
{
if(v == from) continue;
dfs(v, u);
}
}
void solve()
{
cin>>n;
fac[0] = 1;
for(int i = 1; i < N; i++) fac[i] = (fac[i - 1] * i) % mod;
for(int i = 1; i <= n - 1; i++)
{
int u, v; cin>>u>>v;
e[u].push_back(v);
e[v].push_back(u);
}
f[1][1] = dfs1(1, 0);
dfs(1, 0);
for(int i = 1; i <= n; i++)
{
ll sum = 0;
for(int k = 1; k <= n; k++) sum += f[i][k];
sum %= mod;
ll inv = qmi(sum, mod - 2, mod);
for(int k = 1; k <= n; k++)
cout<<(f[i][k] * f[1][1] % mod) * inv % mod<<" ";
cout<<'\n';
}
return;
}
本文来自博客园，作者：magicat，转载请注明原文链接：https://www.cnblogs.com/magicat/p/17706686.html
