本文深入探讨了算法竞赛中的高效解题策略，包括快速幂、动态规划、贪心算法等核心技巧，以及如何利用数据结构如树状数组、哈希表进行优化。通过具体实例分析，展示了如何在比赛中快速准确地解决问题。
n≤2333n\le2333n≤2333
将最后形成的三元组分类
三个一样的，两个一样的，三个都不一样的分别计算就比较方便
三个一样的和两个一样的可以 O(n)O(n)O(n)
三个都不一样显然枚举两个然后算出另一个
关于手写一些可以实现mapmapmap功能的东西
可以用一个类似邻接表的东西存
int first[Mod], nxt[N], to[N], val[N], tot;
int ask(int p){
int x = p % Mod;
for(int i = first[x]; i; i = nxt[i]) if(to[i] == p) return val[i];
return 0;
}
void add(int p, int v){
int x = p % Mod;
for(int i = first[x]; i; i = nxt[i]){ if(to[i] == p) {val[i] += v; return;} }
nxt[++tot] = first[x], first[x] = tot, to[tot] = p, val[tot] = v;
}
也可以冲突就暴力挪位
struct MAP{
static cs int Mod = 1e7+9;
int key[Mod], val[Mod];
Map(){ memset(key, -1, sizeof(key)); }
int pos(int k){
int h = k % Mod;
while(key[h] != -1 && key[h] != k) h = h+1 == Mod ? 0 : h+1;
return h;
}
void ins(int k){
int h = pos(k);
if(key[h] == -1) key[h] = k, val[h] = 1;
else val[h]++;
}
int find(int k){
int h = pos(k);
return key[h] == -1 ? 0 : val[h];
}
};
也可以离散化然后暴力二分
struct MAP{
int b[N], siz, val[N];
void build(){
sort(b + 1, b + siz + 1);
siz = unique(b + 1, b + siz + 1) - (b + 1);
}
int find(int x){
int l = 1, r = siz;
while(l < r){
int mid = (l+r) >> 1;
if(b[mid] >= x) r = mid; else l = mid + 1;
} return l;
}
void add(int x){ val[find(x)]++; }
int ask(int x){ int l = find(x); if(b[l] == x) return val[l]; return 0; }
};
然后发现快速幂的复杂度是满的 log(1e9)log(1e9)log(1e9)，可以改成常数更小的 exgcdexgcdexgcd
还可以预处理逆元，然后 n2n^2n2 枚举的时候就不用快速幂只剩下一个常数很小的二分
复杂度 O(n2log(n))O(n^2log(n))O(n2log(n))
#include<bits/stdc++.h>
#define cs const
using namespace std;
namespace IO{
inline char gc(){
static cs int Rlen = 1 << 22 | 1;
static char buf[Rlen], *p1, *p2;
return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, Rlen, stdin), p1 == p2) ? EOF : *p1++;
}
int read(){
int cnt = 0, f = 1; char ch = 0;
while(!isdigit(ch)) { ch = gc(); if(ch == '-') f = -1;}
while(isdigit(ch)) cnt = cnt * 10 + (ch-'0'), ch = gc();
return cnt * f;
}
}
using namespace IO;
typedef long long ll;
cs int N = 2500;
int n, p, a[N], b[N], siz, inv[N]; ll ans;
int add(int a, int b){ return a + b >= p ? a + b - p : a + b; }
int mul(int a, int b){ return 1ll * a * b % p; }
int ksm(int a, int b){ int ans=1; for(;b;b>>=1, a=mul(a,a)) if(b&1) ans=mul(ans,a); return ans; }
bool vis[N]; int ps[N], tim[N];
struct MAP{
int b[N], siz;
int val[N];
void build(){
sort(b + 1, b + siz + 1);
siz = unique(b + 1, b + siz + 1) - (b + 1);
}
int find(int x){
int l = 1, r = siz;
while(l < r){
int mid = (l+r) >> 1;
if(b[mid] >= x) r = mid; else l = mid + 1;
} return l;
}
void add(int x){ val[find(x)]++; }
int ask(int x){ int l = find(x); if(b[l] == x) return val[l]; return 0; }
}bin, sum;
int main(){
n = read(), p = read();
for(int i = 1; i <= n; i++){
a[i] = read();
bin.b[++bin.siz] = a[i];
sum.b[++sum.siz] = a[i] % p;
b[++siz] = a[i];
inv[i] = ksm(a[i] % p, p - 2);
}
sort(b + 1, b + siz + 1);
siz = unique(b + 1, b + siz + 1) - (b + 1);
bin.build();
sum.build();
for(int i = 1; i <= n; i++)
ps[i] = bin.find(a[i]);
for(int i = 1; i <= n; i++){
if(!tim[ps[i]]) sum.add(a[i] % p);
tim[ps[i]]++;
}
// three of them are the same
for(int i = 1; i <= n; i++){
if(tim[ps[i]] >= 3 && !vis[ps[i]]){
vis[ps[i]] = true;
if(mul(inv[i], mul(inv[i], inv[i])) == 1) ++ans;
}
}
// two of them are the same
memset(vis, 0, sizeof(vis));
for(int i = 1; i <= n; i++){
if(tim[ps[i]] >= 2 && !vis[ps[i]]){
vis[ps[i]] = true;
int nx = mul(inv[i], inv[i]);
ans += sum.ask(nx);
if(a[i] % p == nx) ans--;
}
}
// each of them is different
memset(vis, 0, sizeof(vis));
ll ret = 0;
for(int i = 1; i <= siz; i++) inv[i] = ksm(b[i] % p, p - 2);
for(int i = 1; i <= siz; i++){
for(int j = i + 1; j <= siz; j++){
int ni = b[i] % p, nj = b[j] % p;
int nx = mul(inv[i], inv[j]);
ret += sum.ask(nx);
if(ni == nx) ret--;
if(nj == nx) ret--;
}
} cout << ans + ret / 3; return 0;
}
n≤1e5n\le 1e5n≤1e5
考场的垃圾做法：
发现物品按 www 排序就是选一个 vvv 的最长不下降序列并且满足选出来的每一个 www 在另一边都有一个 ≥\ge≥ 它的映射，另一边也排个序，fi,jf_{i,j}fi,j 表示这边选到 iii另一边选到 jjj 的最长长度
fi,j=max(fp,k)(vp≤vi,k≤j−1)f_{i,j}=max(f_{p,k})(v_p\le v_i,k\le j-1)fi,j=max(fp,k)(vp≤vi,k≤j−1)，二维树状数组优化转移
正解：考虑贪心，显然如果最后的长度是 lenlenlen 的话选最大的 lenlenlen 个一定是最优的
考虑对每一个 iii 求出它向后的最长长度 lenlenlen 并且求出可以装下 iii 的个数 cntcntcnt，那么合法的最长长度就是 min(len,cnt)min(len,cnt)min(len,cnt)，从后向前 dpdpdp，动态维护装的下的个数，树状数组优化转移即可
有时候不能莽起 dp，还是要从贪心的角度考虑
#include<bits/stdc++.h>
#define cs const
using namespace std;
int read(){
int x = 0, f = 1; char c = 0;
while(!isdigit(c)){ c = getchar(); if(c == '-')f = -1;}
while(isdigit(c)) x = x*10 + (c-'0'), c = getchar();
return x * f;
}
typedef long long ll;
cs int N = 1e5 + 5;
int T, n, m, t[N], b[N], siz;
struct BAG{ int w, v;}a[N];
bool cmp(BAG a, BAG b){ return a.w < b.w || (a.w == b.w && a.v < b.v); }
struct BIT{
int c[N]; void clear(){ memset(c, 0, sizeof(c)); }
void add(int x, int v){ for(;x;x-=x&-x) c[x] = max(c[x], v); }
int ask(int x){ int ans=0; for(;x<=siz;x+=x&-x) ans = max(ans, c[x]); return ans; }
}bit;
void Solve(){
bit.clear(); siz = 0;
n = read();
for(int i = 1; i <= n; i++){
a[i].w = read(); a[i].v = read();
b[++siz] = a[i].v;
} sort(a + 1, a + n + 1, cmp);
reverse(a + 1, a + n + 1);
sort(b + 1, b + siz + 1); siz = unique(b + 1, b + siz + 1) - (b + 1);
m = read();
for(int i = 1; i <= m; i++){
t[i] = read();
} sort(t + 1, t + m + 1);
reverse(t + 1, t + m + 1);
int ans = 0, p = 0;
for(int i = 1; i <= n; i++){
while(p + 1 <= m && t[p + 1] >= a[i].w) ++p;
int x = lower_bound(b + 1, b + siz + 1, a[i].v) - b;
int f = min(bit.ask(x) + 1, p);
bit.add(x, f); ans = max(ans, f);
} cout << ans << '\n';
}
int main(){
T = read(); while(T--) Solve(); return 0;
}
n≤500n\le 500n≤500
考场的垃圾做法：
fsiz,depf_{siz,dep}fsiz,dep 表示大小和深度为 siz,depsiz,depsiz,dep 的方案数，暴力枚举拆分，以及拆分出来的深度
fsiz,dep=∑i=1siz−1∑j=1siz−i−1fsiz−i−1,j∑k=1ifi,k[max(j,k+1)=1]f_{siz,dep}=\sum_{i=1}^{siz-1}\sum_{j=1}^{siz-i-1}f_{siz-i-1,j}\sum_{k=1}^{i}f_{i,k}[max(j,k+1)=1]fsiz,dep=∑i=1siz−1∑j=1siz−i−1fsiz−i−1,j∑k=1ifi,k[max(j,k+1)=1]
然后还要重新分配编号，考场上思维不严谨的我写的 (siz−1k)\binom{siz-1}{k}(ksiz−1)
请睁大眼睛看一下我的预处理
c[0][0] = 1;
for(int i = 1; i <= n; i++){ c[i][i] = 1;
for(int j = 1; j <= i; j++){
c[i][j] = add(c[i-1][j], c[i-1][j-1]);
}
}
手玩一下发现这样预处理正好把杨辉三角挪了一位，所以实际上乘的系数是 (siz−2k−1)\binom{siz-2}{k-1}(k−1siz−2)
而 stdstdstd 正好成的是 (siz−2k−1)\binom{siz-2}{k-1}(k−1siz−2)，然后我就歪打正着了（雾）
言归正传，钦定是原来的取到 depdepdep 还是后来拼接的取到 depdepdep 就可以前缀和优化转移
说一下乘 (siz−1k)\binom{siz-1}{k}(ksiz−1) 为什么有问题，因为当前拼接的子树很有可能是跟之前的同构，显然会算重
我们直接钦定当前拼接上去的子树的根是次小，这样相当于对儿子进行了排序，这样就不会算重了
既然钦定了根是最小，当前子树的根是次小，那么方案数就是将剩下的重新排列的方案数即 (siz−2k−1)\binom{siz-2}{k-1}(k−1siz−2)
然后发现前缀和非常丑，直接按套路把状态定义为大小为 sizsizsiz 深度 ≤dep\le dep≤dep 的方案数
fsiz,dep=∑i=1siz−1fsiz−i−1,dep∗fi,dep−1∗(siz−2i−1)f_{siz,dep}=\sum_{i=1}^{siz-1}f_{siz-i-1,dep}*f_{i,dep-1}*\binom{siz-2}{i-1}fsiz,dep=i=1∑siz−1fsiz−i−1,dep∗fi,dep−1∗(i−1siz−2)
考虑直接枚举子树的拼接，枚举子树个数以及每个子树的大小
f[d][n]=∑k=1n−11k!∑∑i=1kbi=n−1(n−1)!∏bi!∏i=1kf[d−1][bi]f[d][n]=\sum_{k=1}^{n-1}\frac{1}{k!}\sum_{\sum_{i=1}^k b_i=n-1}\frac{(n-1)!}{\prod b_i!}\prod_{i=1}^kf[d-1][b_i]f[d][n]=k=1∑n−1k!1∑i=1kbi=n−1∑∏bi!(n−1)!i=1∏kf[d−1][bi]
f[d][n](n−1)!=∑k=1n−11k!∑∑i=1kbi=n−11∏bi!∏i=1kf[d−1][bi]\frac{f[d][n]}{(n-1)!}=\sum_{k=1}^{n-1}\frac{1}{k!}\sum_{\sum_{i=1}^k b_i=n-1}\frac{1}{\prod b_i!}\prod_{i=1}^kf[d-1][b_i](n−1)!f[d][n]=k=1∑n−1k!1∑i=1kbi=n−1∑∏bi!1i=1∏kf[d−1][bi]
注意到后面一坨的组合意义是集合间带标号的拼接，上指数型生成函数
令 fd(x)=∑i=0∞f[d][i]i!xif_d(x)=\sum_{i=0}^{\infty}\frac{f[d][i]}{i!}x^ifd(x)=∑i=0∞i!f[d][i]xi
∑n=0∞f[d][n](n−1)!xn−1=∑n=0∞∑k=1n−11k!∑∑i=1kbi=n−1∏i=1kf[d−1][bi]bi!xbi\sum_{n=0}^{\infty}\frac{f[d][n]}{(n-1)!}x^{n-1}=\sum_{n=0}^{\infty}\sum_{k=1}^{n-1}\frac{1}{k!}\sum_{\sum_{i=1}^k b_i=n-1}\prod_{i=1}^k\frac{f[d-1][b_i]}{b_i!}x^{b_i}n=0∑∞(n−1)!f[d][n]xn−1=n=0∑∞k=1∑n−1k!1∑i=1kbi=n−1∑i=1∏kbi!f[d−1][bi]xbi
=∑k=1∞1k!∑bi=0∞∏i=1kf[d−1][bi]bi!xbi=\sum_{k=1}^{\infty} \frac{1}{k!}\sum_{b_i=0}^{\infty}\prod_{i=1}^k\frac{f[d-1][b_i]}{b_i!}x^{b_i}=k=1∑∞k!1bi=0∑∞i=1∏kbi!f[d−1][bi]xbi
=∑k=1∞1k!∏i=1k∑bi=0∞f[d−1][bi]bi!xbi=\sum_{k=1}^{\infty} \frac{1}{k!}\prod_{i=1}^k\sum_{b_i=0}^{\infty}\frac{f[d-1][b_i]}{b_i!}x^{b_i}=k=1∑∞k!1i=1∏kbi=0∑∞bi!f[d−1][bi]xbi
=∑k=1∞1k!fd−1(x)k=exp(fd−1(x))=\sum_{k=1}^{\infty} \frac{1}{k!}f_{d-1}(x)^k=exp(f_{d-1}(x))=k=1∑∞k!1fd−1(x)k=exp(fd−1(x))
而
∑n=0∞f[d][n](n−1)!xn−1=fd(x)′=exp(fd−1(x))\sum_{n=0}^{\infty}\frac{f[d][n]}{(n-1)!}x^{n-1}=f_d(x)'=exp(f_{d-1}(x))n=0∑∞(n−1)!f[d][n]xn−1=fd(x)′=exp(fd−1(x))
所以
fd(x)=∫exp(fd−1(x))dxf_d(x)=\int exp(f_{d-1}(x)) dxfd(x)=∫exp(fd−1(x))dx
多项式全家桶
#include<bits/stdc++.h>
#define cs const
using namespace std;
int read(){
int x = 0, f = 1; char c = 0;
while(!isdigit(c)){ c = getchar(); if(c == '-')f = -1;}
while(isdigit(c)) x = x*10 + (c-'0'), c = getchar();
return x * f;
}
typedef long long ll;
cs int N = 505;
int n, k, L, R;
cs int Mod = 998244353;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a, b); }
int c[N][N];
bool ban[N];
int f[N][N];
int main(){
n = read();
k = read();
for(int i = 1; i <= k; i++) ban[read()] = true;
L = read();
R = read();
c[0][0] = 1;
for(int i = 1; i <= n; i++){
c[i][0] = 1;
for(int j = 1; j <= i; j++){
c[i][j] = add(c[i-1][j], c[i-1][j-1]);
}
}
f[1][1] = ban[1] ? 0 : 1;
for(int dep = 2; dep <= n; dep++){
for(int siz = 1; siz <= n; siz++){
if(siz == 1){ f[siz][dep] = 1; continue; }
for(int k = 1; k < siz; k++){
Add(f[siz][dep], mul(c[siz - 2][k - 1], mul(f[siz - k][dep], f[k][dep - 1])));
}
} for(int siz = 1; siz <= n; siz++) if(ban[siz]) f[siz][dep] = 0;
}
for(int i = L; i <= R; i++) cout << add(f[n][i], Mod - f[n][i-1]) << " ";
return 0;
}
#include<bits/stdc++.h>
#define cs const
using namespace std;
int read(){
int x = 0, f = 1; char c = 0;
while(!isdigit(c)){ c = getchar(); if(c == '-')f = -1;}
while(isdigit(c)) x = x*10 + (c-'0'), c = getchar();
return x * f;
}
typedef long long ll;
cs int N = 505;
int n, k, L, R;
cs int Mod = 998244353;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a, b); }
int ksm(int a, int b){ int ans = 1; for(;b;b>>=1, a = mul(a, a)) if(b&1) ans = mul(ans, a); return ans; }
int a[N];
#define poly vector<int>
poly f[N];
cs int K = 13;
poly w[K + 1];
int inv[1 << K | 5];
int bit, up, rev[1 << K | 5];
void prework(){
for(int i = 1; i <= K; i++) w[i].resize(1 << i - 1);
int wn = ksm(3, (Mod-1)/(1<<K)); w[K][0] = 1;
for(int i = 1; i < (1 << K-1); i++) w[K][i] = mul(w[K][i-1], wn);
for(int i = K-1; i; i--)
for(int j = 0; j < (1 << i-1); j++) w[i][j] = w[i + 1][j << 1];
inv[0] = inv[1] = 1;
for(int i = 2; i <= (1 << K); i++) inv[i] = mul(Mod-Mod/i, inv[Mod%i]);
}
void init(int len){
bit = 0; up = 1; while(up < len) up <<= 1, bit++;
for(int i = 0; i < up; i++) rev[i] = (rev[i>>1]>>1) | ((i&1)<<(bit-1));
}
void NTT(poly &a, int typ){
for(int i = 0; i < up; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
for(int i = 1, l = 1; i < up; i <<= 1, l++)
for(int j = 0; j < up; j += (i << 1))
for(int k = 0; k < i; k++){
int x = a[k + j], y = mul(w[l][k], a[k + j + i]);
a[k + j] = add(x, y); a[k + j + i] = add(x, Mod - y);
}
if(typ == -1){
reverse(a.begin() + 1, a.end());
for(int i = 0; i < up; i++) a[i] = mul(a[i], inv[up]);
}
}
poly operator * (poly a, poly b){
int len = a.size() + b.size() - 1;
init(len);
a.resize(up); b.resize(up);
NTT(a, 1); NTT(b, 1);
for(int i = 0; i < up; i++) a[i] = mul(a[i], b[i]);
NTT(a, -1); a.resize(len); return a;
}
poly Inv(poly a, int lim){
int n = a.size(); poly c, b(1, ksm(a[0], Mod-2));
for(int len = 4; (len >> 2) < lim; len <<= 1){
init(len); c.resize(len >> 1);
for(int i = 0; i < (len >> 1); i++) c[i] = i < n ? a[i] : 0;
c.resize(up); b.resize(up);
NTT(c, 1); NTT(b, 1);
for(int i = 0; i < up; i++) b[i] = mul(b[i], add(2, Mod - mul(b[i], c[i])));
NTT(b, -1); b.resize(len >> 1);
} b.resize(lim);
return b;
}
poly deriv(poly a){
for(int i = 0; i+1 < a.size(); i++) a[i] = mul(a[i + 1], i + 1);
a.pop_back(); return a;
}
poly integ(poly a){
a.push_back(0);
for(int i = a.size() - 1; i; i--) a[i] = mul(a[i - 1], inv[i]);
a[0] = 0; return a;
}
poly ln(poly a, int len){
a = integ(deriv(a) * Inv(a, len));
a.resize(len); return a;
}
poly exp(poly a, int len){
int n = a.size(); poly c, b(1, 1);
for(int i = 2; (i >> 1) < len; i <<= 1){
c = ln(b, i); c[0] = add(c[0], Mod-1);
for(int j = 0; j < i; j++) c[j] = add(j < n ? a[j] : 0, Mod - c[j]);
b = b * c; b.resize(i);
} b.resize(len); return b;
}
int main(){
prework();
n = read();
k = read();
for(int i = 1; i <= k; i++) a[i] = read();
L = read();
R = read();
f[1].resize(n + 1); f[1][1] = 1; f[0].resize(n + 1);
for(int i = 2; i <= n; i++){
f[i] = integ(exp(f[i - 1], n + 1)); f[i].resize(n + 1);
for(int j = 1; j <= k; j++) f[i][a[j]] = 0;
} int fac = 1;
for(int i = 1; i <= n; i++) fac = mul(fac, i);
for(int i = L; i <= R; i++) cout << mul(fac, add(f[i][n], Mod - f[i - 1][n]))<< " ";
return 0;
}
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