The 3rd Universal Cup. Stage 8: Cangqian

题解：

https://files.cnblogs.com/files/clrs97/ZJCPC24_Tutorial.pdf

Code：

A. Bingo

#include <bits/stdc++.h>
using namespace std;
string n;
int m;
typedef long long ll;
ll sum[1000005];
int pw[1000005];
bool all[1000005];
int sol[1000020];
void solv() {
    cin >> n >> m;
    int am = 0 ;
    for(int i = 0;i < n.size();i++) am = (10LL * am + n[i] - '0') % m;
    int ans = m - am;
     
    sum[n.size()] = 0;
    for(int i = 0;i < n.size();i++) {
        sum[i] = 1LL * ('9' - n[i]) * pw[n.size() - i - 1] ;
    }
    all[n.size()] = 1;
    for(int i = n.size() - 1;i >= 0;i--) {
        all[i] = all[i + 1] && (n[i] == '9') ;
    }
    for(int i = n.size() - 1;i >= 0;i--) sum[i] += sum[i + 1];
    string b; int d = m;
    while(d) {
        b += ((d % 10) + '0') ;
        d /= 10;
    }
    reverse(b.begin() , b.end()) ;
    vector<int> rm(b.size() + 1) ;
    for(int i = b.size() - 1;i >= 0;i--) {
        rm[i] = (b[i] - '0') * pw[b.size() - i - 1] + rm[i + 1] ;
    }
    for(int i = 0;i < n.size() && i + b.size() - 1 < n.size();i++) { //match
        for(int j = 0 ; j <= b.size();j++) {
            /// here mat , diff on n[i + j]
            if(i + j >= n.size()) break ;
            if(j == b.size()) {
                if(!all[i + j]) ans = min(ans , 1) ;
                break;
            }
            if(b[j] > n[i + j]) {
                /// 
                // printf("in %d %d : %d %d %d\n",i,j,(ll)(b[j] - n[i + j] - 1) * pw[n.size() - (i + j) - 1] , sum[i + j + 1] , 1LL * rm[j + 1] * pw[n.size() - (i+b.size())]) ;
                ans = min((ll)ans , 1 + (ll)(b[j] - n[i + j] - 1) * pw[n.size() - (i + j) - 1] + sum[i + j + 1] + 1LL * rm[j + 1] * pw[n.size() - (i+b.size())]);
            }
            if(j == b.size() || i + j >= n.size() || b[j] != n[i + j]) break;
        }
    }
    int L = (int)n.size() - 1;
    // printf("L %d\n",L) ;
    for(int i = 0;i < n.size() + 15;i++) sol[i] = 0;
    for(int i = 0;i < n.size();i++) sol[n.size() - i - 1] = n[i] - '0' ;
    sol[0] += ans ;
    for(int i = 0;i <= L;i++) {
        sol[i + 1] += sol[i] / 10;
        sol[i] %= 10;
        if(sol[i + 1]) L = max(L , i + 1) ;
    }
    for(int i = L;i >= 0;i--) cout << sol[i] ; cout << '\n';
    return ;
}
int main() {
    // freopen("in.txt","r",stdin);
    // freopen("out2.txt","w",stdout) ;
    ios::sync_with_stdio(false) ; cin.tie(0) ; cout.tie(0) ;
    int t;cin >> t;
    pw[0] = 1;
    for(int i = 1;i <= 100000;i++) {
        pw[i] = min(1000000000LL , pw[i - 1] * 10LL) ;
    }
    while(t--) solv() ;
}

B. Simulated Universe

#include<bits/stdc++.h>
using namespace std;
  
const int N=8e3+1e2+7;
  
int T,n;
  
int f[2][N];
  
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>T;
    while(T--)
    {
        cin>>n;
        for(int i=0;i<=n;i++)
            f[0][i]=-1e9;
        f[0][0]=0;
        int now=0,last=1;
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            swap(now,last);
            for(int j=0;j<=i+1;j++)
                f[now][j]=-1e9;
            string ty;
            cin>>ty;
            if(ty=="B")
            {
                ans+=2;
                for(int j=0;j<=i;j++)
                {
                    if(f[last][j]<0)
                        continue;
                    f[now][j+1]=max(f[now][j],f[last][j]);
                    if(f[last][j])
                        f[now][j]=max(f[now][j],f[last][j]-1);
                }
            }
            else
            {
                int a,b;
                cin>>a>>b;
                for(int j=0;j<=i;j++)
                {
                    if(f[last][j]<0)
                        continue;
                    f[now][max(j-a,0)]=max(f[now][max(j-a,0)],f[last][j]);
                    f[now][j]=max(f[now][j],f[last][j]+b);
                }
            }
        }
        for(int i=0;i<=n;i++)
            if(f[now][i]>=0)
            {
                ans-=i;
                break;
            }
        cout<<ans<<"\n";
    }
}
/*
1
3 6
1 2 3
4 5 6
6 9 9
8 12 13
10 15 17
12 18 21
*/

C. Challenge NPC

#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int> pii ;
vector<pii> Ed;
int main() {
    ios::sync_with_stdio(false) ; cin.tie(0) ;
    int k ; cin >> k;
    int n = k + 2;
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j < i;j++) {
            Ed.push_back({j*2 , i*2 - 1});
            Ed.push_back({j*2 - 1 , i *2});
        }
    }
    cout << n*2 << ' ' << Ed.size() <<' ' << 2 << '\n' ;
    for(int i = 1;i <= n*2;i++) {
        cout << (i&1) + 1 <<' ' ;
    }
    cout << '\n';
    for(auto [x,y] : Ed) cout << x <<' ' << y << '\n';
    return 0;
}

D. Puzzle: Easy as Scrabble

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const int N=300005,P=998244353;
#define NO {puts("NO"); exit(0);}
int dx[]={0,0,1,-1},dy[]={1,-1,0,0}; 
auto solve(){
    int n,m;
    cin>>n>>m;
    vector<string> a(n+2);
    for(auto &s:a) cin>>s;
    vector f(n+2,vector<array<char,4>>(m+1));
    vector inq(n+2,vector<bool>(m+2));
    for(int i=1;i<=n;i++){
        if(a[i][0]!='.')
            f[i][1][0]=a[i][0];
        if(a[i][m+1]!='.')
            f[i][m][1]=a[i][m+1];
    }
    for(int j=1;j<=m;j++){
        if(a[0][j]!='.')
            f[1][j][2]=a[0][j];
        if(a[n+1][j]!='.')
            f[n][j][3]=a[n+1][j];
    }
    auto check=[&](int x,int y){
        static int vis[256];
        static int time;
        int num=0; ++time;
        for(auto c:f[x][y]){
            if(!isalpha(c)) continue;
            if(a[x][y]!='x') a[x][y]=c;
            if(vis[c]!=time){
                vis[c]=time;
                num++;
            }
        }
        if(num>1) return true;
        return num==1&&a[x][y]=='x';
    };
    queue<pair<int,int>> qu;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(check(i,j)){
                qu.push({i,j});
                inq[i][j]=true;
            }
        }
    }
    while(qu.size()){
        auto [x,y]=qu.front(); qu.pop();
        inq[x][y]=false;
        a[x][y]='x';
        for(int i=0;i<4;i++){
            if(!isalpha(f[x][y][i])) continue;
            int nx=x+dx[i],ny=y+dy[i];
            if(nx<=0||ny<=0||nx>n||ny>m) NO;
            f[nx][ny][i]=f[x][y][i];
            if(a[nx][ny]!='x') a[nx][ny]=f[x][y][i];
            if(!inq[nx][ny]&&check(nx,ny)){
                inq[nx][ny]=true;
                qu.push({nx,ny});
            }
        }
        f[x][y]={0,0,0,0};
    }
    puts("YES");
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            if(a[i][j]=='x') a[i][j]='.';
            putchar(a[i][j]);
        }
        puts("");
    }
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    // cin>>t;
    while(t--){
        solve();
        // cout<<solve()<<'\n';
        // cout<<(solve()?"Yes":"No")<<'\n';
    }
    return 0;
}
  
  
/* Generated by powerful Codeforces Tool(cf tool)
 * Author: sleep__
 * Time: 2024-04-04 14:25:05
**/

E. Team Arrangement

#include<cstdio>
#include<algorithm>
using namespace std;
typedef unsigned long long ull;
const int N=65,inf=~0U>>1;
int n,i,w[N],num[N],ans;ull in[N],out[N];
struct E{int l,r;}e[N];
inline bool cmp(const E&a,const E&b){return a.r<b.r;}
inline void solve(int m){
  int sum=0;
  ull S=0;
  for(int i=1;i<=m;i++){
    S^=in[i];
    int now=num[i];
    sum+=now*w[i];
    now*=i;
    while(now--){
      if(!S)return;
      S-=S&-S;
    }
    S^=S&out[i];
  }
  if(sum>ans)ans=sum;
}
void dfs(int x,int m){
  if(x>m)return;
  num[x]=0;
  dfs(x+1,m);
  for(int i=1;;i++){
    m-=x;
    num[x]=i;
    if(!m)solve(x);
    if(m<=0)break;
    dfs(x+1,m);
  }
}
int main(){
  scanf("%d",&n);
  for(i=0;i<n;i++)scanf("%d%d",&e[i].l,&e[i].r);
  sort(e,e+n,cmp);
  for(i=0;i<n;i++){
    in[e[i].l]^=1ULL<<i;
    out[e[i].r]^=1ULL<<i;
  }
  for(i=1;i<=n;i++)scanf("%d",&w[i]);
  ans=-inf;
  dfs(1,n);
  if(ans==-inf)puts("impossible");else printf("%d",ans);
}

F. Stage: Agausscrab

#include<bits/stdc++.h>
using namespace std;
  
const int N=1e3+1e2+7;
  
int n;
  
string s[N];
  
int a[N];
  
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>s[i]>>a[i];
    string ans;
    for(int i=1;i<=n;i++)
    {
        int r=0;
        for(int j=1;j<=n;j++)
            if(a[j]>a[i])
                r++;
        r+=1;
        for(int j=1;j<=r;j++)
            if(s[i].size())
                s[i].pop_back();
        ans+=s[i];
    }
    ans[0]=ans[0]-'a'+'A';
    cout<<"Stage: "<<ans<<"\n";
}

G. Crawling on a Tree

#include<cstdio>
#include<algorithm>
#include<cstdlib>
using namespace std;
typedef long long ll;
const int N=10005,M=10005;
int n,m,i,lim[N],g[N],v[N<<1],w[N<<1],wt[N<<1],nxt[N<<1],ed;
int wf[N],wk[N],need[N],sz[N],heavy[N];
ll ans[M];
struct E{
  ll s,d[M];
  int l,r;
  void clr(){
    s=l=0;r=m;
    for(int i=0;i<=m;i++)d[i]=0;
  }
}f[15];
inline void add(int x,int y,int z,int k){
  v[++ed]=y;
  w[ed]=z;
  wt[ed]=k;
  nxt[ed]=g[x];
  g[x]=ed;
}
inline void merge(const E&A,const E&B,E&C){
  int l=A.l+B.l,r=min(A.r+B.r,m);
  static ll d[M];
  int i=A.l+1,j=B.l+1,k=l+1;
  while(k<=r&&i<=A.r&&j<=B.r)d[k++]=A.d[i]<B.d[j]?A.d[i++]:B.d[j++];
  while(k<=r&&i<=A.r)d[k++]=A.d[i++];
  while(k<=r&&j<=B.r)d[k++]=B.d[j++];
  C.s=A.s+B.s;
  C.l=l;
  C.r=r;
  for(int i=l+1;i<=r;i++)C.d[i]=d[i];
}
void dfs(int x,int y){
  need[x]=lim[x];
  sz[x]=1;
  for(int i=g[x];i;i=nxt[i]){
    int u=v[i];
    if(u==y)continue;
    wf[u]=w[i];
    wk[u]=wt[i];
    dfs(u,x);
    sz[x]+=sz[u];
    if(sz[u]>sz[heavy[x]])heavy[x]=u;
    need[x]=max(need[x],need[u]);
  }
}
void go(int x,int y,int o){
  int u=heavy[x];
  if(u){
    go(u,x,o);
    f[o+1].clr();
    merge(f[o],f[o+1],f[o]);
  }else f[o].clr();
  for(int i=g[x];i;i=nxt[i]){
    int u=v[i];
    if(u==y||u==heavy[x])continue;
    go(u,x,o+1);
    merge(f[o],f[o+1],f[o]);
  }
  int L=need[x],R=wk[x],W=wf[x];
  int A=max(f[o].l,L*2-R),B=min(f[o].r,R);
  if(A>B){
    for(int i=1;i<=m;i++)puts("-1");
    exit(0);
  }
  for(int i=f[o].l+1;i<=A;i++)f[o].s+=f[o].d[i];
  f[o].l=A;
  f[o].r=B;
  f[o].s+=1LL*(max(L,A)*2-A)*W;
  for(int i=A+1;i<=B&&i<=L;i++)f[o].d[i]-=W;
  for(int i=max(A,L)+1;i<=B;i++)f[o].d[i]+=W;
}
int main(){
  scanf("%d%d",&n,&m);
  for(i=1;i<n;i++){
    int x,y,z,k;
    scanf("%d%d%d%d",&x,&y,&z,&k);
    add(x,y,z,k);
    add(y,x,z,k);
  }
  for(i=2;i<=n;i++)scanf("%d",&lim[i]);
  wk[1]=m*2;
  dfs(1,0);
  go(1,0,0);
  for(i=1;i<=m;i++)ans[i]=-1;
  ll sum=f[0].s;
  for(i=f[0].l;i<=m;i++){
    if(i>=need[1])ans[i]=sum;
    if(i<m)sum+=f[0].d[i+1];
  }
  for(i=1;i<=m;i++)printf("%lld\n",ans[i]);
}

H. Permutation

#include <bits/stdc++.h>
using namespace std;
int n;
const double C = (sqrt(5) - 1) / 2;
int f[1000005];
int d[1000005];
int qry(int l,int r) {
    cout << "? " << l <<' ' << r << '\n' ;
    fflush(stdout) ;
    int x ; cin >> x;
    return x;
}
void solv() {
    cin >> n;
    int l = 1 , r = n;
    int lst_pos = -1;
    while(l < r) {
        int c = d[r - l + 1];
        // printf("C %d %d %d\n",l,r,c) ;
        if(lst_pos == -1) lst_pos = qry(l , r);
        if(r - l == 1) {
            if(lst_pos == l) l = r;
            else r = l;
            break ;
        }
        if(l + c - 1 >= lst_pos) {
            int x = qry(l , l + c - 1);
            if(x == lst_pos) r = l + c - 1;
            else {
                l = l + c;
                lst_pos = -1;
            }
        }
        else {
            int x = qry(r - c + 1 , r);
            if(x == lst_pos) l = r - c + 1;
            else {
                r = r - c;
                lst_pos = -1;
            }
        }
    }
    cout << "! " << l << '\n' ;
    fflush(stdout) ;
    return ;
}
int main() {
    // ios::sync_with_stdio(false) ; cin.tie(0) ;
    int t;cin >> t;
    f[1] = f[2] = 0;
    for(int i = 3;i <= 1000000;i++) {
        int mxc = (int)(C * i); 
        f[i] = 1e9;
        for(int j = max(mxc - 10 , (i + 1)/2) ; j <= min(i - 1 , mxc + 10) ; j++) {
            if(max(f[j] + 1 , f[i - j] + 2) < f[i]) {
                f[i] = max(f[j] + 1 , f[i - j] + 2) ;
                d[i] = j;
            }
        }
        // if(f[i] > ceil(1.5 * log2(i)) - 1) {
        //     printf("%d %d %lf\n",i,f[i],ceil(1.5 * log2(i)) - 1);
        // }
    }
    // printf("%d\n",f[1000000]) ;?
    while(t--) solv() ;
}

I. Piggy Sort

100

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102

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106

107

108

109

110

111

112

#include<bits/stdc++.h>
using namespace std;
  
#define int long long
  
const int N=2e3+1e2+7;
  
int T,n,m;
  
int x[N][N],sx[N];
  
map<int,int> vis[N];
  
int use[N];
  
int av[N],bv[N],fd,ans[N];
  
void dfs(int t)
{
    if(t==n+1)
    {
        vector<int> id(n);
        iota(id.begin(),id.end(),1);
        sort(id.begin(),id.end(),[&](const int &a,const int &b){
            if(av[a]*bv[b]!=av[b]*bv[a])
                return av[a]*bv[b]<av[b]*bv[a];
            return a<b;
        });
        fd=1;
        for(int i=0;i<n;i++)
            ans[id[i]]=i+1;
        for(int i=1;i<=n;i++)
            cout<<ans[i]<<" \n"[i==n];
        return;
    }
    for(int i=1;i<=n;i++)
    {
        if(use[i])
            continue;
        int va=x[2][i]-x[1][t];
        int vb=sx[2]-sx[1];
        int ok=1;
        for(int j=3;j<=m;j++)
        {
            if(va*(sx[j]-sx[1])%vb)
            {
                ok=0;
                break;
            }
            int w=va*(sx[j]-sx[1])/vb+x[1][t];
            if(!vis[j].count(w)||!vis[j][w])
            {
                ok=0;
                break;
            }
        }
        if(ok)
        {
            for(int j=1;j<=m;j++)
            {
                int w=va*(sx[j]-sx[1])/vb+x[1][t];
                vis[j][w]--;
            }
            use[i]=1;
            av[t]=va,bv[t]=vb;
            dfs(t+1);
            use[i]=0;
            for(int j=1;j<=m;j++)
            {
                int w=va*(sx[j]-sx[1])/vb;
                vis[j][w]++;
            }
        }
    }
}
  
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(int i=1;i<=m;i++)
        {
            vis[i].clear();
            sx[i]=0;
            for(int j=1;j<=n;j++)
                cin>>x[i][j],vis[i][x[i][j]]++,sx[i]+=x[i][j];
        }
        if(sx[1]==sx[2])
        {
            for(int i=1;i<=n;i++)
                cout<<i<<" \n"[i==n];
            continue;
        }
        fd=0;
        dfs(1);
        assert(fd);
    }
}
/*
1
3 6
1 2 3
4 5 6
6 9 9
8 12 13
10 15 17
12 18 21
*/

J. Even or Odd Spanning Tree

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107

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=200005,K=18,M=500005,inf=~0U>>1;
int Case,n,m,cnt,i,j,x,y,z,tmp,f[N],g[N],v[N<<1],w[N<<1],nxt[N<<1],ed;
int d[N],fa[N][K],fe[N][K],fo[N][K];
long long mst,ans;int delta;bool on[M];
struct E{int x,y,z;}e[M];
inline bool cmp(const E&a,const E&b){return a.z<b.z;}
inline void umax(int&a,int b){a<b?(a=b):0;}
int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
inline void add(int x,int y,int z){v[++ed]=y;w[ed]=z;nxt[ed]=g[x];g[x]=ed;}
void dfs(int x,int y){
  for(int i=1;i<K;i++){
    fa[x][i]=fa[fa[x][i-1]][i-1];
    fe[x][i]=max(fe[x][i-1],fe[fa[x][i-1]][i-1]);
    fo[x][i]=max(fo[x][i-1],fo[fa[x][i-1]][i-1]);
  }
  for(int i=g[x];i;i=nxt[i]){
    int u=v[i],z=w[i];
    if(u==y)continue;
    d[u]=d[x]+1;
    fa[u][0]=x;
    if(z&1){
      fe[u][0]=0;
      fo[u][0]=z;
    }else{
      fe[u][0]=z;
      fo[u][0]=0;
    }
    dfs(u,x);
  }
}
inline int ask(int x,int y,int w[][K]){
  if(x==y)return 0;
  if(d[x]<d[y])swap(x,y);
  int ret=0;
  for(int i=K-1;~i;i--)if(d[fa[x][i]]>=d[y]){
    umax(ret,w[x][i]);
    x=fa[x][i];
  }
  if(x==y)return ret;
    for(int i=K-1;~i;i--)if(fa[x][i]!=fa[y][i]){
    umax(ret,w[x][i]);
    umax(ret,w[y][i]);
    x=fa[x][i];
    y=fa[y][i];
  }
  umax(ret,w[x][0]);
  umax(ret,w[y][0]);
  return ret;
}
int main(){
  scanf("%d",&Case);
  while(Case--){
    scanf("%d%d",&n,&m);
    for(i=0;i<=n;i++){
      g[i]=d[i]=0;
      for(j=0;j<K;j++)fa[i][j]=fe[i][j]=fo[i][j]=0;
    }
    mst=cnt=ed=0;
    for(i=1;i<=m;i++){
      scanf("%d%d%d",&e[i].x,&e[i].y,&e[i].z);
      on[i]=0;
    }
    sort(e+1,e+m+1,cmp);
    for(i=1;i<=n;i++)f[i]=i;
    for(i=1;i<=m;i++){
      x=e[i].x,y=e[i].y,z=e[i].z;
      if(F(x)==F(y))continue;
      on[i]=1;
      f[f[x]]=f[y];
      mst+=z;
      cnt++;
      add(x,y,z),add(y,x,z);
    }
    if(cnt<n-1){
      puts("-1 -1");
      continue;
    }
    dfs(1,0);
    delta=inf;
    for(i=1;i<=m;i++){
      if(on[i])continue;
      z=e[i].z;
      tmp=ask(e[i].x,e[i].y,z&1?fe:fo);
      if(!tmp)continue;
      z-=tmp;
      if(delta>z)delta=z;
    }
    if(delta<inf)ans=mst+delta;else ans=-1;
    if(mst&1)swap(ans,mst);
    printf("%lld %lld\n",mst,ans);
  }
}
/*
3
2 1
1 2 5
3 1
1 3 1
4 4
1 2 1
1 3 1
1 4 1
2 4 2
*/

K. Sugar Sweet 3

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#include <bits/stdc++.h>
using namespace std;
int A , B , C  , x;
int n ;
const int N = 605;
const int mod = 1e9 + 7;
int fpow(int a,int b) {
    int ans = 1;
    while(b) {
        if(b & 1) ans = 1LL * ans * a %mod;
        a = 1LL * a * a %mod; b>>= 1;
    }
    return ans;
}
int F[N][N];
// int Ga[N][N] , Gb[N][N], Gc[N][N];
// int dota[N][N] , dotb[N][N] , dotc[N][N];
int dot[N][N];
int Cata[N] ;
int d[N] ;
int t[N * 2] , rt[N * 2] ;
int lm ; ///多项式的项数
int mat[N][N] , rmat[N][N];
int C_(int a,int b) {
    return 1LL * t[a] * rt[b] % mod * rt[a - b] % mod;
}
int A_(int a,int b) {
    return 1LL * t[a] * rt[a - b] % mod;
}
vector<int> get_coe(int n , vector<int> w) { ///已知n个点值[1 to n]，求sum{ai * wi} 对应的 di的系数，i from 0 to n - 1
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= n;j++) {
            mat[i][j] = fpow(i , j - 1) ; // MAT * A = D
            rmat[i][j] = (i == j) ;
        }
    }
    for(int i = 1;i <= n;i++) {
        int cur = -1;
        for(int j = i ;j <= n;j++) {
            if(mat[j][i]) {cur = j ; break ;}
        }
        for(int j = 1;j <= n;j++) {swap(mat[cur][j] , mat[i][j]) ; swap(rmat[cur][j] , rmat[i][j]) ;}
        int r = fpow(mat[i][i] , mod - 2);
        for(int j = 1;j <= n;j++) {
            if(i == j) continue ;
            int f = (mod - 1LL * mat[j][i] * r %mod) % mod;
            for(int k = 1;k <= n;k++) {mat[j][k] = (mat[j][k] + 1LL * mat[i][k] * f) % mod ; rmat[j][k] = (rmat[j][k] + 1LL * rmat[i][k] * f) % mod;}
        }
        for(int j = 1;j <= n;j++) {
            mat[i][j] = 1LL * mat[i][j] * r % mod;
            rmat[i][j] = 1LL * rmat[i][j] * r % mod;
        }
    }
    // for(int i = 1;i <= n;i++ , printf("\n")) for(int j = 1;j <= n;j++) printf("%d ",rmat[i][j]) ;
    ///A = Rmat * D
     
    vector<int> coe(n) ;
    for(int i = 0;i < n;i++) {
        for(int j = 0;j < n;j++) {
            coe[j] = (coe[j] + 1LL * w[i] * rmat[i + 1][j + 1]) % mod;
        }
    }
    return coe;
}
  
int main() {
    cin >> A >> B >> C >> x;
    n = A + B + C;
    if(n & 1) {
        cout << 0 ; return 0;
    }
    if(A > n/2 || B > n/2 || C > n/2) {
        cout << 0 ; return 0;
    }
    t[0] = rt[0] = 1;
    for(int i = 1;i <= n + 1;i++) t[i] = 1LL * t[i - 1] * i % mod , rt[i] = fpow(t[i] , mod - 2) ;
    Cata[0] = 1;
    for(int i = 1;i <= n/2;i++) Cata[i] = 1LL * C_(i*2 , i) * fpow(i + 1 , mod - 2) % mod;
  
    //// F[i][j]表示i个主元分j段的方案数，Fi(x)作为其egf
    F[0][0] = 1;
    lm = n / 2 + 1;
    for(int i = 1;i <= n/2;i++) {
        for(int j = 1;j <= i;j++) {
            for(int k = 1;k <= i;k++) { ///最后一段包含的主元个数
                F[i][j] = (F[i][j] + 1LL * F[i - k][j - 1] * Cata[k - 1]) % mod;
            }
            // printf("%d %d : %d\n",i,j,F[i][j]) ;
        }
    }
    for(int i = 0;i <= n/2;i++) {
        for(int j = 0;j <= i;j++) F[i][j] = 1LL * F[i][j] * rt[j] % mod ; ///变成egf
        for(int j = 1;j <= lm;j++) {
            dot[i][j] = 0;
            for(int k = i;k >= 0;k--) dot[i][j] = (1LL * dot[i][j] * j + F[i][k]) % mod;
            ///dot 代表点值
        }
    }
    vector<int> w(n/2 + 1) ;
    for(int i = 0;i <= n/2;i++) w[i] = 1LL * fpow(i , x) * t[i] % mod;
     
    vector<int> coe = get_coe(n/2 + 1 , w) ; /// coe.len = n/2 + 1
     
     
    int ans = 0;
    for(int a = 0;a <= n/2 && a <= A;a++) {
        for(int b = 0;b + a <= n / 2 && b <= B;b++) {
            int c = n / 2 - a - b;
            if(c > C) continue ;
            /// F[a] * F[b] * F[c]
            for(int i = 1;i <= lm;i++) d[i] = 1LL * dot[a][i] * dot[b][i] % mod * dot[c][i] % mod;
            // for(int i = 1;i <= lm;i++) printf("%d ",3LL * d[i] % mod) ; printf("\n") ;
            int sol = 0;
            for(int i = 1;i <= lm;i++) sol = (sol + 1LL * d[i] * coe[i - 1]) % mod;
  
  
            int sol2 = 0;
            for(int ab = 0;ab <= A - a && ab <= b; ab++){
                int ac = (A - a - ab) ;
                int bc = (c - ac) ;
                int ba = (B - b - bc) ;
                int ca = (a - ba) ;
                int cb = (C - c - ca) ;
                if(ac < 0 || bc < 0 || ba < 0 || ca < 0 || cb < 0) continue ;
                sol2 = (sol2 + 1LL * C_(a , ba) * C_(b , ab) % mod * C_(c , ac)) % mod;
            }
            // printf("%d %d %d : %d %d\n",a,b,c,sol,sol2);
  
            ans = (ans + 1LL * sol * sol2) % mod;
        }
    }
    cout << ans << '\n';
}

L. Challenge Matrix Multiplication

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#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
typedef pair<int,int> pii;
int to[N] , fir[N] , nxt[N] ;
bool ok[N];
int cc = 0;
int nc , nodes[N];
void adde(int u,int v) {
    ++cc;
    to[cc] = v;
    nxt[cc] = fir[u];
    fir[u] = cc;
}
  
int in[N] , out[N] ;
int n ,m ;
  
  
int ans[N];
  
  
bool vis[N];
pii fr[N] ;
int cnt = 0;
int qu[N] , l , r;
void bfs(int u) {
    l = 1 , r = 0;
    qu[++r] = u;
    vis[u] = 1;
    while(l <= r) {
        int u = qu[l++];
        cnt++;
        for(int i = fir[u] ; i ; i = nxt[i]) {
            if(!vis[to[i]]) {
                qu[++r] = to[i] ;
                vis[to[i]] = 1;
            }
        }
    }
    return ;
}
int main() {
    ios::sync_with_stdio(false) ; cin.tie(0); cout.tie(0);
    clock_t cl = clock() ;
    cin >>n>>m; //printf("??? %d %d\n",n,m);
     
    for(int i = 1;i <= m;i++) {
        int u , v;cin >> u >> v;
        in[v]++ ; out[u]++;
        adde(u , v);
    }
    for(int i = 1;i <= n;i++) ans[i] = 1;
    int all = m;
    clock_t sum0 = 0 , sum1 = 0;
    while(all) {
        // clock_t a = clock();
        for(int i = 1;i <= n;i++) {
            if(in[i] < out[i]) {
                memset(vis,0,sizeof(vis)) ;
                vis[i] = 1;
                int lst;
                for(int j = i ; j <= n;j++) {
                    if(!vis[j]) continue ;
                    if(in[j] > out[j]) {lst = j ; break ;}
                    for(int x = fir[j] ; x ; x = nxt[x]) {
                        if(!vis[to[x]] && !ok[x]) {
                            vis[to[x]] = 1;
                            fr[to[x]] = {j , x};
                        }
                    }
                }
                 
                int u = lst ;
                nc = 0;
                while(u) {
                    nodes[nc++] = u;
                    if(u != i) in[u]--;
                    if(u != lst) out[u]--;
                    if(u == i) break ;
                    ok[fr[u].second] = 1;
                    u = fr[u].first ;
                    all--;
                }
                break ;
            }
        }
        // sum0 += (clock() - a);
        // a = clock() ;
        cnt = 0;
        memset(vis,0,sizeof(vis));
        for(int i = 0;i < nc;i++) {
            bfs(nodes[i]) ;
            ans[nodes[i]] = cnt;
        }
        // static int gg = 0;
        // sum1 += (clock() - a) ;
        // printf("i-th round %d %d\n",++gg , nc) ;
    }
    // cerr << (double)(clock() - cl) / CLOCKS_PER_SEC <<' ' << (double)sum0 / CLOCKS_PER_SEC <<' ' << (double)sum1 / CLOCKS_PER_SEC << '\n';
    // return 0;
    for(int i = 1;i <= n;i++) cout << ans[i] <<' ' ;
    cout << '\n';
}

M. Triangles

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios_base::sync_with_stdio(false);
    long long n,a,b;
    cin>>n>>a>>b;
    long long ans=0;
    for(long long r=1;r<=n;r++)
    {
        ans+=r*(r+1);
        if(r*2>n)ans-=(r*2-n)*(r*2-n+1)/2;
    }
    for(long long i=1;i<=b;i++)
    {
        ans-=(a-b+1);
        ans-=max(min(a+1-i,n-a)-(b-i),0ll);
    }
    cout<<ans<<endl;
    return 0;
}