今天又考的奇差
题解：
第一题；
这么简单一道题我想了好久，智商实在是下线了；
#include<bits/stdc++.h> using namespace std; int main(){ freopen("shortway.in","r",stdin); freopen("shortway.out","w",stdout); int n, k; scanf("%d%d", &n, &k); if((n - 1) % k == 0)printf("%d ", (n - 1) / k * 2); if((n - 1) % k == 1)printf("%d ", (n - 1) / k * 2 + 1); if((n - 1) % k >= 2)printf("%d ", (n - 1) / k * 2 + 2); }
第二题：我们发现e的长度很小，我们可以在上面做文章，其实每个位置对应的%strlen(e)都是一样的；我们用树状数组维护rt[pos][len][yu][id]表示1到POS每次跳len，当前位置%len = yu的id字符有多少个；对于一个字符串，查询就是rt[R][len][(L + i - 1) % len][a[i]],修改就把长度枚举一下；
这道题我开始想到每次跳len, 维护一个前缀和，但没有想到树状数组，暴力修改和我直接暴力复杂度差不多；如果想到余数这一维就好了
#include<bits/stdc++.h> using namespace std; const int M = 1e5 + 10; char s[M], ch[M]; int rt[M][11][10][4], n, a[M], b[M]; inline int get(char x){ switch(x){ case 'A':return 0; case 'T':return 1; case 'C':return 2; case 'G':return 3; } } inline int lowbit(int x){return x & (-x);} void add(int pos, int len, int yu, int id, int d){ for(int x = pos; x <= n; x += lowbit(x)){ rt[x][len][yu][id] += d; } } int query(int pos, int id, int k, int yu){ int ret = 0; for(int x = pos; x; x -= lowbit(x)){ ret += rt[x][k][yu][id]; } return ret; } void init(){ for(int i = 1; i <= n; i++) for(int k = 1; k <= 10; k++){ add(i, k, i % k, a[i], +1); } } void add(int pos, int id, int d){ for(int k = 1; k <= 10; k++) add(pos, k, pos % k, id, d); } int main(){ freopen("evolution.in","r",stdin); freopen("evolution.out","w",stdout); scanf("%s", s); n = strlen(s); for(int i = 1; i <= n; i++)a[i] = get(s[i - 1]); init(); int q, opt, l, r; scanf("%d", &q); while(q--){ scanf("%d", &opt); if(opt == 1){ scanf("%d%s", &l, ch); int t = get(ch[0]); add(l, a[l], -1); add(l, t, +1); a[l] = t; } else { scanf("%d%d%s", &l, &r, ch ); int ans = 0, len = strlen(ch); for(int i = 1; i <= len; i++){ b[i] = get(ch[i - 1]); ans += query(r, b[i], len, (l + i - 1) % len); ans -= query(l - 1, b[i], len, (l + i - 1) % len); } printf("%d ", ans); } } }
第三题：
线段树存树边，维护最大值更新非树遍，用非树边更新线段树上的限制；
#include<bits/stdc++.h> using namespace std; const int M = 2e5 + 10; const int oo = 0x3f3f3f3f; #define ex(i, u) for(int i = h[u]; i; i = G[i].nxt) #define Ls nd->ls, lf, mid #define Rs nd->rs, mid + 1, rg int fa[M], ans[M], n, m, h[M], top[M], siz[M], son[M], dep[M], in[M]; int idc, tot, seq[M], line[M]; bool intree[M], fg; struct edge{int u, v, w, id;}g[M]; struct Edge{int nxt, v, id;}G[M << 1]; void add(int u, int v, int i){ G[++tot].v = v; G[tot].nxt = h[u]; h[u] = tot; G[tot].id = g[i].id; } bool cmp(edge a, edge b){ return a.w < b.w; } bool cmp_id(edge a, edge b){ return a.id < b.id; } int find(int x){ if(x == fa[x])return x; return fa[x] = find(fa[x]); } void uni(int u, int v){ fa[find(u)] = find(v); } struct Node; void modify(int L, int R, int d, Node * nd, int lf, int rg); struct Node{ Node *ls, *rs; int mx, mi, tag; inline void down(int lf, int rg){ if(tag != oo){ int mid = (lf + rg) >> 1; modify(lf, mid, tag, ls, lf, mid); modify(mid + 1, rg, tag, rs, mid + 1, rg); tag = oo; } } inline void up(){ mx = max(ls->mx, rs->mx); } }pool[M << 2], *tail = pool, *root; Node * build(int lf = 1, int rg = n){ Node *nd = ++tail; if(lf == rg) { nd->mi = nd->tag = oo; nd->mx = g[line[seq[lf]]].w; } else { int mid = (lf + rg) >> 1; nd->ls = build(lf, mid); nd->rs = build(mid + 1, rg); nd->tag = nd->mi = oo; nd->up(); } return nd; } int query(int L, int R, Node * nd = root, int lf = 1, int rg = n){ if(L <= lf && rg <= R) return nd->mx; else { nd->down(lf, rg); int mid = (lf + rg) >> 1; int ans = 0; if(L <= mid)ans = max(ans, query(L, R, Ls)); if(R > mid)ans = max(ans, query(L, R, Rs)); return ans; } } void modify(int L, int R, int d, Node * nd = root, int lf = 1, int rg = n){ if(L <= lf && rg <= R) { nd->tag = min(nd->tag, d); nd->mi = min(nd->mi, d); return ; } else { nd->down(lf, rg); int mid = (lf + rg) >> 1; if(L <= mid)modify(L, R, d, Ls); if(R > mid)modify(L, R, d, Rs); //nd->up();// ??? } } void getdown(Node * nd = root, int lf = 1, int rg = n){ if(lf == rg){ ans[line[seq[lf]]] = nd->mi; return ; } nd->down(lf, rg); int mid = (lf + rg) >> 1; getdown(Ls); getdown(Rs); } void dfs(int u, int f){ dep[u] = dep[f] + 1; siz[u] = 1; fa[u] = f; ex(i, u){ int v = G[i].v; if(v == f)continue; dfs(v, u); line[v] = G[i].id; siz[u] += siz[v]; if(siz[v] > siz[son[u]])son[u] = v; } } void dfs2(int u, int tp){ in[u] = ++idc; seq[idc] = u; top[u] = tp; if(son[u])dfs2(son[u], tp); ex(i, u){ int v = G[i].v; if(v == fa[u] || v == son[u])continue; dfs2(v, v); } } void Modify(int u, int v, int d){ while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]])swap(u, v); modify(in[top[u]], in[u], d); u = fa[top[u]]; } if(dep[u] < dep[v])swap(u, v); if(in[u] != in[v])modify(in[v] + 1, in[u], d); } int Query(int u, int v){ int ans = 0; while(top[u] != top[v]){ if(dep[top[u]] < dep[top[v]])swap(u, v); ans = max(ans, query(in[top[u]], in[u])); //if(fg)printf("%d %d %d %d %d %d %d ", u, v, dep[top[u]], dep[top[v]], in[top[u]], in[u], ans); u = fa[top[u]]; } if(dep[u] < dep[v])swap(u, v); if(in[u] != in[v])ans = max(ans, query(in[v] + 1, in[u])); //if(fg)printf("%d %d %d ", u, v, ans); return ans; } int main(){ freopen("mst.in","r",stdin); freopen("mst.out","w",stdout); int u, v, w; scanf("%d%d", &n, &m); for(int i = 1; i <= m; i++){ scanf("%d%d%d", &u, &v, &w); g[i] = (edge){u, v, w, i}; } sort(g + 1, g + 1 + m, cmp); for(int i = 1; i <= n; i++)fa[i] = i; for(int i = 1; i <= m; i++){ int u = g[i].u, v = g[i].v; if(find(u) == find(v))continue; uni(u, v); intree[g[i].id] = 1; add(u, v, i); add(v, u, i); } line[1] = 0; dfs(1, 0); dfs2(1, 1); //for(int i = 1; i <= n; i++)printf("%d %d %d ", i, dep[i], top[i]);puts(""); sort(g + 1, g + 1 + m, cmp_id); root = build(); for(int i = 1; i <= m; i++){ if(intree[i])continue; int u = g[i].u, v = g[i].v; if(i == 473)fg = 1; int t = Query(u, v); ans[i] = t - 1; Modify(u, v, g[i].w - 1); fg = 0; } getdown(); for(int i = 1; i <= m; i++) printf("%d ",ans[i] == oo ? -1 : ans[i]); }