「PKUWC2018/PKUSC2018」試題選做
「PKUWC2018/PKUSC2018」試題選做
最近還沒想好報THUSC還是PKUSC,THU發我的三類約(再來一瓶)不知道要不要用,甚至不知道營還辦不辦,協議還有沒有用。所以這些事情就暫時先不管了,PKU的題還是不錯的,就刷一刷劃水。因為比較簡單,所以就不單獨寫博客了。
loj2537 Minimax
數據結構題,兩個 \(\log\) 直接啟發式合並,一個 \(\log\) 需要轉移的時候多維護一些東西。
對於每個節點維護一下選擇其子樹裏每個葉子的權值的概率,線段樹合並轉移即可。
code
/*program by mangoyang*/ #include <bits/stdc++.h> #define inf (0x7f7f7f7f) #define Max(a, b) ((a) > (b) ? (a) : (b)) #define Min(a, b) ((a) < (b) ? (a) : (b)) typedef long long ll; using namespace std; template <class T> inline void read(T &x){ int ch = 0, f = 0; x = 0; for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1; for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48; if(f) x = -x; } const int M = 1e9, N = 300005, mod = 998244353; int a[N], rt[N], P, n, ans; vector<int> g[N]; vector<pair<int, int> > Ans; namespace Seg{ #define mid ((l + r) >> 1) int s[N*30], tag[N*30], lc[N*30], rc[N*30], size; inline void init(){ for(int i = 1; i < N * 30; i++) tag[i] = 1; } inline void pushdown(int u){ if(tag[u] == 1) return; if(lc[u]){ s[lc[u]] = 1ll * tag[u] * s[lc[u]] % mod; tag[lc[u]] = 1ll * tag[lc[u]] * tag[u] % mod; } if(rc[u]){ s[rc[u]] = 1ll * tag[u] * s[rc[u]] % mod; tag[rc[u]] = 1ll * tag[rc[u]] * tag[u] % mod; } tag[u] = 1; } inline void insert(int &u, int l, int r, int pos){ if(!u) u = ++size; if(l == r) return (void) (s[u] = 1); if(pos <= mid) insert(lc[u], l, mid, pos); else insert(rc[u], mid + 1, r, pos); s[u] = s[lc[u]] + s[rc[u]]; } inline int merge(int x, int y, int sumx, int sumy){ if(!x){ if(!sumx) return y; s[y] = 1ll * s[y] * sumx % mod; tag[y] = 1ll * tag[y] * sumx % mod; return y; } if(!y){ if(!sumy) return x; s[x] = 1ll * s[x] * sumy % mod; tag[x] = 1ll * tag[x] * sumy % mod; return x; } pushdown(x), pushdown(y); int o = ++size; int Lx = (sumx + 1ll * P * s[lc[x]] % mod) % mod; int Ly = (sumy + 1ll * P * s[lc[y]] % mod) % mod; int Rx = (sumx + 1ll * (mod + 1 - P) * s[rc[x]] % mod) % mod; int Ry = (sumy + 1ll * (mod + 1 - P) * s[rc[y]] % mod) % mod; lc[o] = merge(lc[x], lc[y], Rx, Ry); rc[o] = merge(rc[x], rc[y], Lx, Ly); s[o] = (s[lc[o]] + s[rc[o]]) % mod; return o; } inline void getans(int u, int l, int r){ if(l == r) return (void) Ans.push_back(make_pair(l, s[u])); pushdown(u); if(lc[u]) getans(lc[u], l, mid); if(rc[u]) getans(rc[u], mid + 1, r); } } inline int Pow(int a, int b){ int ans = 1; for(; b; b >>= 1, a = 1ll * a * a % mod) if(b & 1) ans = 1ll * ans * a % mod; return ans; } inline void dfs(int u){ if(!(int) g[u].size()) Seg::insert(rt[u], 1, M, a[u]); for(int i = 0; i < (int) g[u].size(); i++){ dfs(g[u][i]); P = a[u]; rt[u] = Seg::merge(rt[u], rt[g[u][i]], 0, 0); } } int main(){ read(n); for(int i = 1, x; i <= n; i++) read(x), g[x].push_back(i); for(int i = 1; i <= n; i++){ read(a[i]); if((int) g[i].size()) a[i] = 1ll * a[i] * Pow(10000, mod1- 2) % mod; } Seg::init(); dfs(1), Seg::getans(rt[1], 1, M), sort(Ans.begin(), Ans.end()); for(int i = 0; i < (int) Ans.size(); i++){ (ans += 1ll * (i + 1) * Ans[i].first % mod * Ans[i].second % mod * Ans[i].second % mod) %= mod; } cout << ans << endl; return 0; }
loj2538 Slay the Spire
一個顯然的貪心策略是從大到小盡可能打出強化牌(至多 \(k-1\) 張),然後從大到小盡可能打出攻擊牌。
可以分別對兩種牌排序然後按照這個策略設計 dp。
\(f[i][j]\) 表示前 \(i\) 張強化牌摸了 \(j\) 張的期望加成,從大到小對強化牌排序,枚舉當前這張牌是否被摸到來轉移。
\(g[i][j]\) 表示前 \(i\) 張攻擊牌摸了 \(j\) 張的期望攻擊力之和,\(j\) 越大可以打的攻擊牌越多,所以從小到大排序來轉移。
code
/*program by mangoyang*/ #include <bits/stdc++.h> #define inf (0x7f7f7f7f) #define Max(a, b) ((a) > (b) ? (a) : (b)) #define Min(a, b) ((a) < (b) ? (a) : (b)) typedef long long ll; using namespace std; template <class T> inline void read(T &x){ int ch = 0, f = 0; x = 0; for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1; for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48; if(f) x = -x; } const int N = 3005, mod = 998244353; int a[N], b[N], js[N], inv[N], f[2][N], g[2][N], n, m, k; inline void up(int &x, int y){ x = (x + y >= mod ? x + y - mod : x + y); } inline int Pow(int a, int b){ int ans = 1; for(; b; b >>= 1, a = 1ll * a * a % mod) if(b & 1) ans = 1ll * ans * a % mod; return ans; } inline int C(int x, int y){ return 1ll * js[x] * inv[y] % mod * inv[x-y] % mod; } inline void solve(){ read(n), read(m), read(k); for(int i = 1; i <= n; i++) read(a[i]); for(int i = 1; i <= n; i++) read(b[i]); sort(a + 1, a + n + 1, greater<int>()); sort(b + 1, b + n + 1); for(int i = 0; i <= n; i++) f[0][i] = f[1][i] = g[0][i] = g[1][i] = 0; f[0][0] = 1; for(int i = 1, o = 1; i <= n; i++, o ^= 1){ for(int j = 0; j <= i; j++){ f[o][j] = f[o^1][j]; if(j && j < k) up(f[o][j], 1ll * a[i] * f[o^1][j-1] % mod); else if(j) up(f[o][j], f[o^1][j-1]); } } for(int i = 1, o = 1; i <= n; i++, o ^= 1){ for(int j = 0; j <= i; j++){ g[o][j] = g[o^1][j]; if(j){ if(m - j < k - 1) up(g[o][j], g[o^1][j-1]); up(g[o][j], 1ll * C(i - 1, j - 1) * b[i] % mod); } } } int ans = 0; for(int i = 0; i <= min(m, n); i++) if(m - i <= n) up(ans, 1ll * f[n&1][i] * g[n&1][m-i] % mod); printf("%d\n", ans); } int main(){ js[0] = inv[0] = 1; for(int i = 1; i < N; i++) js[i] = 1ll * js[i-1] * i % mod, inv[i] = Pow(js[i], mod - 2); int T; read(T); while(T--) solve(); return 0; }
loj2540 隨機算法
直接狀壓 \(dp\) 是 \(O(3^nn)\) 的,實際上選了會沖突的點與已經選了的點顯然可以看做同一類點,然後就 \(O(2^nn)\) 了。
code
/*program by mangoyang*/ #include<bits/stdc++.h> #define inf (0x7f7f7f7f) #define Max(a, b) ((a) > (b) ? (a) : (b)) #define Min(a, b) ((a) < (b) ? (a) : (b)) typedef long long ll; using namespace std; template <class T> inline void read(T &x){ int ch = 0, f = 0; x = 0; for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1; for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48; if(f) x = -x; } const int mod = 998244353; int cnt[1<<21], dp[1<<21][21], js[21], inv[21], ed[21], n, m; inline int Pow(int a, int b){ int ans = 1; for(; b; b >>= 1, a = 1ll * a * a % mod) if(b & 1) ans = 1ll * ans * a % mod; return ans; } inline int A(int x, int y){ return 1ll * js[x] * inv[x-y] % mod; } inline void up(int &x, int y){ x = (x + y >= mod ? x + y - mod : x + y); } int main(){ read(n), read(m); js[0] = inv[0] = 1; for(int i = 1; i <= n; i++){ js[i] = 1ll * js[i-1] * i % mod; inv[i] = Pow(js[i], mod - 2); } for(int i = 1, x, y; i <= m; i++){ read(x), read(y); ed[x] |= 1 << (y - 1); ed[y] |= 1 << (x - 1); } for(int s = 0; s < (1 << n); s++) for(int i = 0; i < n; i++) if((1 << i) & s) cnt[s]++; dp[0][0] = 1; for(int s = 0; s < (1 << n) - 1; s++) for(int i = 0; i < n; i++) if(dp[s][i]){ for(int j = 0; j < n; j++) if(!((1 << j) & s)){ int t = s | (1 << j) | ed[j+1]; up(dp[t][i+1], 1ll * dp[s][i] * A(n - cnt[s] - 1, cnt[s^t] - 1) % mod); } } int mx = 0; for(int i = 1; i <= n; i++) if(dp[(1<<n)-1][i]) mx = i; cout << 1ll * dp[(1<<n)-1][mx] * inv[n] % mod << endl; return 0; }
loj2542 隨機遊走
設 \(dp[x][s]\) 為從 \(x\) 出發走到點集 \(s\) 中任意一個點的期望步數,然後 \(\min-\max\) 容斥即可 ,直接分層圖+高斯消元跑不過去,可以用小套路把未知數設出來以後每個點都可以用其父親的答案乘上一個系數再加上一個常數來表示,然後跑一遍樹形dp就做完了。
code
/*program by mangoyang*/
#include<bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 1 << 21, mod = 998244353;
vector<int> g[N];
int mn[N], cnt[N], in[N], A[N], B[N], n, q, rt;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline void dfs(int u, int fa){
if(in[u]) return (void) (A[u] = B[u] = 0);
int SA = 0, SB = 0, deg = (int) g[u].size();
for(int i = 0; i < (int) g[u].size(); i++){
int v = g[u][i];
if(v == fa) continue;
dfs(v, u);
SA = (SA + A[v]) % mod, SB = (SB + B[v]) % mod;
}
deg = Pow(deg, mod - 2);
int x = (1 - 1ll * SA * deg % mod + mod) % mod;
x = Pow(x, mod - 2);
A[u] = 1ll * deg * x % mod;
B[u] = 1ll * x * (1ll * SB * deg % mod + 1) % mod;
}
int main(){
read(n), read(q), read(rt);
for(int i = 1, x, y; i < n; i++){
read(x), read(y);
g[x].push_back(y), g[y].push_back(x);
}
for(int s = 0; s < (1 << n); s++){
for(int i = 1; i <= n; i++){
in[i] = ((1 << (i - 1)) & s) ? 1 : 0;
cnt[s] += in[i];
}
dfs(rt, 0), mn[s] = B[rt];
}
while(q--){
int num, sta = 0;
read(num);
for(int i = 1, x; i <= num; i++) read(x), sta |= 1 << (x - 1);
int ans = 0;
for(int s = sta; s; s = (s - 1) & sta)
(ans += 1ll * ((cnt[s] & 1) ? 1 : mod - 1) * mn[s] % mod) %= mod;
printf("%d\n", ans);
}
return 0;
}
loj6432 真實排名
對於每個點枚舉一下要不要,統計一下會影響的人數,組合數算一下就好了
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
#define fi first
#define se second
const int N = 1000005, mod = 998244353;
pair<int, int> a[N];
int js[N], inv[N], ans[N], n, k;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % mod)
if(b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline int C(int x, int y){
if(x < y || x < 0 || y < 0) return 0;
return 1ll * js[x] * inv[y] % mod * inv[x-y] % mod;
}
int main(){
read(n), read(k);
js[0] = inv[0] = 1;
for(int i = 1; i <= n; i++){
js[i] = 1ll * js[i-1] * i % mod;
inv[i] = Pow(js[i], mod - 2);
read(a[i].fi), a[i].se = i;
}
sort(a + 1, a + n + 1);
int p = 0, q = 0;
for(int i = 1; i <= n; i++){
if(a[i].fi == a[i-1].fi && i > 1){
ans[a[i].se] = ans[a[i-1].se];
continue;
}
while(a[p+1].fi * 2 < a[i].fi && p < n) p++;
while(a[q+1].fi < a[i].fi * 2 && q < n) q++;
int tot1 = i - p - 1, tot2 = max(q, i) - i + 1;
(ans[a[i].se] += C(n - tot1 - 1, k)) %= mod;
(ans[a[i].se] += C(n - tot2, k - tot2)) %= mod;
}
for(int i = 1; i <= n; i++) printf("%d\n", ans[i]);
return 0;
}
loj6433 最大前綴和
考慮每一個集合成為最大前綴和的方案數,這個等於其內部排列滿足全部選是最大前綴和乘上剩余元素無論怎麽排都不存在一個前綴和 \(>0\) 的方案數,兩部分分開來狀壓 \(dp\) 即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = (1 << 20), mod = 998244353;
int a[N], f[N], sum[N], g[N], n;
inline void up(int &x, int y){
x = x + y >= mod ? x + y - mod : x + y;
}
int main(){
read(n);
for(int i = 1; i <= n; i++) read(a[i]);
for(int s = 1; s < (1 << n); s++)
for(int i = 0; i < n; i++)
if((1 << i) & s) sum[s] += a[i+1];
g[0] = 1;
for(int i = 0; i < n; i++) f[1<<i] = 1;
for(int s = 0; s < (1 << n); s++)
for(int i = 0; i < n; i++) if(!((1 << i) & s)){
int t = s ^ (1 << i);
if(sum[s] > 0) up(f[t], f[s]);
if(sum[t] <= 0) up(g[t], g[s]);
}
int ans = 0;
for(int s = 0; s < (1 << n); s++){
sum[s] = (sum[s] % mod + mod) % mod;
up(ans, 1ll * sum[s] * f[s] % mod * g[((1<<n)-1)^s] % mod);
}
cout << ans << endl;
return 0;
}
loj6436 星際穿越
可以發現每個點出發最多會向右走一步,於是倍增出每個點走 \(2^j\) 步能走到的最左端點,已經走到這之間所有點的距離和,每次詢問拆成兩個以 \(x\) 結尾的後綴來詢問即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 300005;
ll a[N], f[N][21], s[N][21];
int n, q;
inline ll calc(int x, int L){
if(a[x] <= L) return x - L;
ll ans = x - a[x], now = 1; x = a[x];
for(int i = 20; ~i; i--) if(f[x][i] >= L){
ans += s[x][i] + (x - f[x][i]) * now;
now += 1 << i, x = f[x][i];
}
return ans + (x - L) * (now + 1);
}
int main(){
read(n), a[1] = 1;
for(int i = 2; i <= n; i++) read(a[i]);
for(int i = n; i >= 1; i--){
f[i][0] = min(a[i], i == n ? inf : f[i+1][0]);
s[i][0] = i - f[i][0];
}
for(int j = 1; j <= 20; j++)
for(int i = 1; i <= n; i++){
f[i][j] = f[f[i][j-1]][j-1];
s[i][j] = s[i][j-1] + s[f[i][j-1]][j-1];
s[i][j] += (f[i][j-1] - f[f[i][j-1]][j-1]) << (j - 1);
}
read(q);
for(int i = 1, l, r, x; i <= q; i++){
read(l), read(r), read(x);
ll ans = calc(x, l) - calc(x, r + 1);
ll g = __gcd(ans, (ll) r - l + 1);
printf("%lld/%lld\n", ans / g, (r - l + 1) / g);
}
return 0;
}
loj6436 神仙的遊戲
border有關的計數題可以從等差數列和周期兩個角度考慮,這個題顯然不能等差數列於是就直接上周期定理。
套用周期定理得到如果 \(i\) 是一個 border,那麽所有模 \(|S|-i\) 同余的位置必須相同,也就是不能出現兩個距離為 \(|S|-i-1\) 的 \(0,1\) 位置,直接用 FTT 統計所有差的信息即可。
code
/*program by mangoyang*/
#include <bits/stdc++.h>
#define inf (0x7f7f7f7f)
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = (1 << 22), P = 998244353, G = 3;
char s[N];
int a[N], b[N];
namespace poly{
int rev[N], lg, len;
inline int Pow(int a, int b){
int ans = 1;
for(; b; b >>= 1, a = 1ll * a * a % P)
if(b & 1) ans = 1ll * ans * a % P;
return ans;
}
inline void timesinit(int lenth){
for(len = 1, lg = 0; len <= lenth; len <<= 1, lg++);
for(int i = 0; i < len; i++)
rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (lg - 1));
}
inline void dft(int *a, int sgn){
for(int i = 0; i < len; i++)
if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int k = 2; k <= len; k <<= 1){
int w = Pow(G, (P - 1) / k);
if(sgn == - 1) w = Pow(w, P - 2);
for(int i = 0; i < len; i += k){
int now = 1;
for(int j = i; j < i + (k >> 1); j++){
int x = a[j], y = 1ll * a[j+(k>>1)] * now % P;
a[j] = x + y >= P ? x + y - P : x + y;
a[j+(k>>1)] = x - y < 0 ? x - y + P : x - y;
now = 1ll * now * w % P;
}
}
}
if(sgn == -1){
int Inv = Pow(len, P - 2);
for(int i = 0; i < len; i++)
a[i] = 1ll * a[i] * Inv % P;
}
}
}
using poly::timesinit;
using poly::dft;
int main(){
scanf("%s", s); int n = strlen(s);
for(int i = 0; i < n; i++)
a[i] = s[i] == '0', b[i] = s[i] == '1';
reverse(b, b + n);
timesinit(n + n - 1);
dft(a, 1), dft(b, 1);
for(int i = 0; i < poly::len; i++)
a[i] = 1ll * a[i] * b[i] % P;
dft(a, -1);
ll ans = 1ll * n * n;
for(int i = 1; i < n; i++){
int flag = 0;
for(int j = i; j < n; j += i)
if(a[n-1+j] || a[n-1-j]) flag = 1;
if(!flag) ans ^= 1ll * (n - i) * (n - i);
}
cout << ans << endl;
return 0;
}
「PKUWC2018/PKUSC2018」試題選做