CF1101

D:题意：树上每个顶点有个权值，求最长链，满足链上gcd > 1

解：对每个质数建虚树，每个点只会拆成log个点，所以是log²的。

 #include <bits/stdc++.h>

 const int N = ;

 struct Edge {
     int nex, v, len;
 }edge[N << ], EDGE[N << ]; int tp, TP;

 int p[N], top, e[N], pos2[N], num2, ST[N << ][], pw[N << ], n, ans, id[N], last[N];
 int vis[N];
 int imp[N], K, E[N], val[N], fa[N], d[N], stk[N], use[N], Time, RT, f[N];
 std::vector<int> v[N];

 inline void add(int x, int y) {
     tp++;
     edge[tp].v = y;
     edge[tp].nex = e[x];
     e[x] = tp;
     return;
 }

 inline void ADD(int x, int y) {
     TP++;
     EDGE[TP].v = y;
     EDGE[TP].len = d[y] - d[x];
     EDGE[TP].nex = E[x];
     E[x] = TP;
     return;
 }

 inline void getp(int n) {
     for(int i = ; i <= n; i++) {
         if(!vis[i]) {
             p[++top] = i;
             last[i] = i;
             id[i] = top;
         }
         for(int j = ; j <= top && i * p[j] <= n; j++) {
             vis[i * p[j]] = ;
             last[i * p[j]] = p[j];
             if(i % p[j] == ) break;
         }
     }
     return;
 }

 void DFS_1(int x, int f) {
     fa[x] = f;
     d[x] = d[f] + ;
     pos2[x] = ++num2;
     ST[num2][] = x;
     for(int i = e[x]; i; i = edge[i].nex) {
         int y = edge[i].v;
         if(y == f) {
             continue;
         }
         DFS_1(y, x);
         ST[++num2][] = x;
     }
     return;
 }

 inline void prework() {
     for(int i = ; i <= num2; i++) pw[i] = pw[i >> ] + ;
     for(int j = ; j <= pw[num2]; j++) {
         for(int i = ; i + ( << j) -  <= num2; i++) {
             if(d[ST[i][j - ]] < d[ST[i + ( << (j - ))][j - ]])
                 ST[i][j] = ST[i][j - ];
             else
                 ST[i][j] = ST[i + ( << (j - ))][j - ];
         }
     }
     return;
 }

 inline int lca(int x, int y) {
     x = pos2[x];
     y = pos2[y];
     if(x > y) std::swap(x, y);
     int t = pw[y - x + ];
     if(d[ST[x][t]] < d[ST[y - ( << t) + ][t]])
         return ST[x][t];
     else
         return ST[y - ( << t) + ][t];
 }

 inline bool cmp(const int &a, const int &b) {
     return pos2[a] < pos2[b];
 }

 inline void work(int x) {
     if(use[x] != Time) {
         use[x] = Time;
         f[x] = ;
         E[x] = ;
     }
     return;
 }

 inline void build_t() {
     std::sort(imp + , imp + K + , cmp);
     TP = top = ;
     work(imp[]);
     stk[++top] = imp[];
     for(int i = ; i <= K; i++) {
         int x = imp[i], y = lca(x, stk[top]);
         work(x); work(y);
         while(top >  && pos2[y] <= pos2[stk[top - ]]) {
             ADD(stk[top - ], stk[top]);
             top--;
         }
         if(y != stk[top]) {
             ADD(y, stk[top]);
             stk[top] = y;
         }
         stk[++top] = x;
     }
     while(top > ) {
         ADD(stk[top - ], stk[top]);
         top--;
     }
     RT = stk[top];
     return;
 }

 void DFS(int x) {
     int a = , b = ;
     for(int i = E[x]; i; i = EDGE[i].nex) {
         int y = EDGE[i].v;
         DFS(y);
         if(EDGE[i].len > ) continue;
         if(f[y] > a) {
             b = a;
             a = f[y];
         }
         else b = std::max(b, f[y]);
     }
     if(vis[x] == Time) {
         ans = std::max(ans, a + b + );
         f[x] = a + ;
     }
     return;
 }

 int main() {
     getp(N - );
     memset(vis, , sizeof(vis));
     int tot = top, flag = ;

     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d", &val[i]);
         int x = val[i];
         if(x > ) flag = ;
         while(x > ) {
             int y = last[x];
             while(x % y == ) {
                 x /= y;
             }
             v[id[y]].push_back(i);
         }
     }
     for(int i = , x, y; i < n; i++) {
         scanf("%d%d", &x, &y);
         add(x, y); add(y, x);
     }

     if(flag) {
         puts("");
         return ;
     }

     DFS_1(, );
     prework();

     for(int i = ; i <= tot; i++) {
         K = ;
         ++Time;
         for(int j = ; j < (int)v[i].size(); j++) {
             imp[++K] = v[i][j];
             vis[v[i][j]] = Time;
         }
         build_t();
         DFS(RT);
     }

     printf("%d \n", ans);
     return ;
 }

AC代码

G：题意：给定序列，你要把它分成尽可能多的几段，每段的权值是异或和。

要求没有哪些段的权值异或和为0。输出最大段数。无解-1。

解：考虑无解，肯定是总异或和为0。否则一定存在解。

发现这个东西，其实等价于选出一些前缀，因为异或可以抵消，所以这些前缀能表示出的和这些段能表示出的是相同的。

然后就把前缀插入线性基，看能插入多少个。

看到异或就要想线性基。

 #include <bits/stdc++.h>

 const int N = ;

 int b[], a[N];

 inline int insert(int x) {
     for(int i = ; i >=  && x; i--) {
         if(((x >> i) & ) == ) continue;
         if(b[i]) x ^= b[i];
         else {
             b[i] = x;
             return ;
         }
     }
     return ;
 }

 int main() {

     int n;
     scanf("%d", &n);
     for(int i = ; i <= n; i++) {
         scanf("%d", &a[i]);
         a[i] ^= a[i - ];
     }

     if(!a[n]) {
         printf("-1\n");
         return ;
     }

     int cnt = ;
     for(int i = n; i >= ; i--) {
         cnt += insert(a[i]);
     }

     printf("%d \n", cnt);
     return ;
 }

AC代码