A.
考虑把(u,v)的询问离线挂在u上，然后dfs，每次从fath[x]到[x]相当于x子树dis区间加1，x子树以外区间-1，然后维护区间和区间平方和等。
常数略大。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #define maxv 100500 #define maxe 200500 using namespace std; long long n,fath[maxv],q,uu,vv,dis[maxv],l[maxv],r[maxv],times=0,fdfn[maxv],g[maxv],nume=1,ans[maxv],cnt; long long tot=0,root,ls[maxv<<2],rs[maxv<<2],val1[maxv<<2],val2[maxv<<2],lazy[maxv<<2]; vector <long long> v[maxv],id[maxv]; bool vis[maxv]; struct edge { long long v,nxt; }e[maxe]; long long read() { char ch;long long data=0; while (ch<'0' || ch>'9') ch=getchar(); while (ch>='0' && ch<='9') { data=data*10+ch-'0'; ch=getchar(); } return data; } void addedge(long long u,long long v) { e[++nume].v=v;e[nume].nxt=g[u];g[u]=nume; e[++nume].v=u;e[nume].nxt=g[v];g[v]=nume; } void pushup(long long now,long long left,long long right) { val1[now]=val1[ls[now]]+val1[rs[now]]; val2[now]=val2[ls[now]]+val2[rs[now]]; } void pushdown(long long now,long long left,long long right) { if (!lazy[now]) return; long long mid=left+right>>1,d=lazy[now],ll=mid-left+1,rr=right-mid; val1[ls[now]]+=2*d*val2[ls[now]]+d*d*ll;val1[rs[now]]+=2*d*val2[rs[now]]+d*d*rr; val2[ls[now]]+=d*ll;val2[rs[now]]+=d*rr; lazy[ls[now]]+=d;lazy[rs[now]]+=d; lazy[now]=0; } void build(long long &now,long long left,long long right) { now=++tot;lazy[now]=0; if (left==right) { val1[now]=dis[fdfn[left]]*dis[fdfn[left]];val2[now]=dis[fdfn[left]]; return; } long long mid=left+right>>1; build(ls[now],left,mid); build(rs[now],mid+1,right); pushup(now,left,right); } long long ask(long long now,long long left,long long right,long long l,long long r) { pushdown(now,left,right); if ((left==l) && (right==r)) return val1[now]; long long mid=left+right>>1; if (r<=mid) return ask(ls[now],left,mid,l,r); else if (l>=mid+1) return ask(rs[now],mid+1,right,l,r); else return ask(ls[now],left,mid,l,mid)+ask(rs[now],mid+1,right,mid+1,r); } void modify(long long now,long long left,long long right,long long l,long long r,long long val) { if (l>r) return; pushdown(now,left,right); if ((left==l) && (right==r)) { lazy[now]+=val; val1[now]+=2*val*val2[now]+val*val*(right-left+1);val2[now]+=val*(right-left+1); return; } long long mid=left+right>>1; if (r<=mid) modify(ls[now],left,mid,l,r,val); else if (l>=mid+1) modify(rs[now],mid+1,right,l,r,val); else { modify(ls[now],left,mid,l,mid,val); modify(rs[now],mid+1,right,mid+1,r,val); } pushup(now,left,right); } void get_ans(long long x) { for (long long i=0;i<v[x].size();i++) ans[id[x][i]]=ask(root,1,n,l[v[x][i]],r[v[x][i]]); } void modify_tree(long long x,long long f) { modify(root,1,n,l[x],r[x],-f); modify(root,1,n,1,l[x]-1,f); modify(root,1,n,r[x]+1,n,f); } void dfs1(long long x) { l[x]=r[x]=++times;fdfn[times]=x; for (long long i=g[x];i;i=e[i].nxt) { long long v=e[i].v; if (v!=fath[x]) { dis[v]=dis[x]+1; dfs1(v); r[x]=max(r[x],r[v]); } } } void dfs2(long long x) { get_ans(x); for (long long i=g[x];i;i=e[i].nxt) { long long v=e[i].v; if (v!=fath[x]) { modify_tree(v,1); dfs2(v); modify_tree(v,-1); } } } int main() { n=read(); for (long long i=1;i<=n-1;i++) {fath[i+1]=read();addedge(fath[i+1],i+1);} q=read(); for (long long i=1;i<=q;i++) { uu=read();vv=read(); v[uu].push_back(vv);id[uu].push_back(i); } dfs1(1);build(root,1,n); dfs2(1); for (long long i=1;i<=q;i++) printf("%lld ",ans[i]); return 0; }
B.
我们发现k=0的时候可以o(1)计算（毕竟是01序列）。当k>=1的时候可以证明答案是ceil(l/2)*trunc(l/2)。
要善于猜结论啊。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 500500 using namespace std; long long n,q,a[maxn],s1[maxn],s[maxn],x[maxn],y[maxn],k[maxn],cnt[2][maxn],ans[maxn],tab[maxn]; long long read() { char ch;long long data=0; while (ch<'0' || ch>'9') ch=getchar(); while (ch>='0' && ch<='9') { data=data*10+ch-'0'; ch=getchar(); } return data; } void calc(long long type,long long pos) { cnt[0][0]=1;tab[0]=0; for (long long i=1;i<=n;i++) { cnt[0][i]=cnt[0][i-1];cnt[1][i]=cnt[1][i-1]; if (!s[i]) cnt[0][i]++; else cnt[1][i]++; tab[i]=tab[i-1]+cnt[s[i]^1][i]; } if (!type) { for (long long i=1;i<=n;i++) { if (x[i]>=2) ans[i]=(cnt[1][y[i]-1]-cnt[1][x[i]-2])*cnt[0][y[i]]+(cnt[0][y[i]-1]-cnt[0][x[i]-2])*cnt[1][y[i]]-(tab[y[i]-1]-tab[x[i]-2]); else ans[i]=cnt[1][y[i]-1]*cnt[0][y[i]]+cnt[0][y[i]-1]*cnt[1][y[i]]-tab[y[i]-1]; } } } int main() { n=read();q=read(); for (long long i=1;i<=n;i++) {a[i]=read();s1[i]=s1[i-1]^a[i];s[i]=s1[i];} for (long long i=1;i<=q;i++) {x[i]=read();y[i]=read();k[i]=read();x[i]++;y[i]++;} calc(0,0); for (int i=1;i<=q;i++) { if (k[i]) { long long l=y[i]-x[i]+2; ans[i]=(l/2+(l&1))*(l/2); } } for (long long i=1;i<=q;i++) printf("%lld ",ans[i]); return 0; }
C.
这TM的是个环套树啊。、
我们找出路径上的值，然后每次加gcd(环长，m）加到最大即可。
之前10分的原因是神TM没开long long。（我就说我怎么可能写挂这种题）（撤回）
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxv 100500 #define maxe 200500 using namespace std; struct edge { long long v,w,nxt; }e[maxe]; long long n,g[maxv],nume=1,dis1[maxv],dis2[maxv],root,father[maxv],anc[maxv][20],dep[maxv]; long long r1,r2,q,s,t,m,u,v,w,a,b,x; long long read() { char ch;long long data=0; while (ch<'0' || ch>'9') ch=getchar(); while (ch>='0' && ch<='9') { data=data*10+ch-'0'; ch=getchar(); } return data; } void addedge(long long u,long long v,long long w) {e[++nume].v=v;e[nume].w=w;e[nume].nxt=g[u];g[u]=nume;} long long getfather(long long x) { if (x!=father[x]) father[x]=getfather(father[x]); return father[x]; } bool unionn(long long a,long long b) { long long f1=getfather(a),f2=getfather(b); if (f1==f2) return true; father[f1]=f2;return false; } void dfs(long long x,long long father) { for (long long i=g[x];i;i=e[i].nxt) { long long v=e[i].v; if (v!=father) { anc[v][0]=x;dep[v]=dep[x]+1; dis1[v]=dis1[x]+e[i].w;dis2[v]=dis2[x]+e[i^1].w; dfs(v,x); } } } void get_table() { for (long long e=1;e<=19;e++) for (long long i=1;i<=n;i++) anc[i][e]=anc[anc[i][e-1]][e-1]; } long long lca(long long x,long long y) { if (dep[x]<dep[y]) swap(x,y); for (long long e=19;e>=0;e--) if ((dep[anc[x][e]]>=dep[y]) && (anc[x][e])) x=anc[x][e]; if (x==y) return x; for (long long e=19;e>=0;e--) { if (anc[x][e]!=anc[y][e]) { x=anc[x][e]; y=anc[y][e]; } } return anc[x][0]; } long long gcd(long long a,long long b) { if (!b) return a; return gcd(b,a%b); } long long calc(long long x) { x=(x%m+m)%m; long long f1=(r1%m+m)%m,d1=gcd(f1,m); return (m-1-x)/d1*d1+x; } int main() { n=read(); for (long long i=1;i<=n;i++) father[i]=i; for (long long i=1;i<=n;i++) { a=read();b=read();x=read(); if (unionn(a,b)) {root=a;u=a;v=b;w=x;} else {addedge(a,b,x);addedge(b,a,-x);} } dfs(root,0);get_table(); r1=dis1[v]-w;r2=dis2[v]+w; q=read(); for (long long i=1;i<=q;i++) { s=read();t=read();m=read();x=lca(s,t); printf("%lld ",calc(dis2[s]-dis2[x]+dis1[t]-dis1[x])); } return 0; }