前言

$$\begin{gathered} \text{暴力写挂} \\ \text{垫底了呀} \end{gathered}$$

$$\text{\%\%\% ZZY rank1}$$

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成绩

$J$表示初中，$H$表示高中后面加的是几年级

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$1$	$(J-3)ZZY$	$250$	$100$	$50$	$100$
$5$	$(J-3)WHF$	$200$	$0$	$100$	$100$
$21$	$(J-2)HJQ$	$100$	$0$	$100$	$0$
$28$	$(J-2)ZYC$	$80$	$40$	$30$	$10$
$32$	$(J-2)HZB$	$70$	$20$	$50$	$0$
$32$	$(J-3)QYH$	$70$	$0$	$0$	$70$
$37$	$(J-2)LRZ$	$50$	$0$	$50$	$0$
$50$	$(J-2)WYC$	$30$	$20$	$10$	$0$
$50$	$(J-2)LW$	$30$	$0$	$30$	$0$

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正题

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$T1:jzoj3360-[NOI2013$模拟$]$苹果树【树上莫队$,LCA$】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95624497

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$T2:jzoj3362,bzoj3758-[NOI2013$模拟$]$数数【分段打表$,$背包$,$状压】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95645249

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$T3:P4556,jzoj3397-[GDOI2014$模拟$]$雨天的尾巴【树链剖分$,$线段树】

链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/95635820

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$someofcode$

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$T120$分$code$

#pragma GCC optimize(2)
%:pragma GCC optimize(3)
%:pragma GCC optimize("Ofast")
%:pragma GCC optimize("inline")
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=51000,M=110000;
struct Que_node{int l,r,a,b,id,pos;
}q[M];
struct Edge_node{int to,next;
}a[N*2];
int n,m,c[N],v[N],ls[N],tot,num,ans[M];
int dfn[N],cnt,fa[N],dep[N],rfn[N];
bool operator<(Que_node x,Que_node y)
{return x.pos==y.pos?x.l<y.l:x.pos<y.pos;}
void addl(int x,int y)
{a[++tot].to=y;a[tot].next=ls[x];ls[x]=tot;
}
void dfs(int x,int Fa)
{fa[x]=Fa;dfn[++cnt]=x;rfn[x]=cnt;for(int i=ls[x];i;i=a[i].next){int y=a[i].to;if(y==Fa) continue;dep[y]=dep[x]+1;dfs(y,x);}
}
void add(int x)
{if(!v[x]) num++;v[x]++;}
void del(int x)
{v[x]--;if(!v[x]) num--;}
void LCA(int x,int y)
{if(dep[x]>dep[y]) swap(x,y);while(dep[x]!=dep[y]){add(c[y]);y=fa[y];}while(x!=y){add(c[x]);add(c[y]);x=fa[x];y=fa[y];}add(c[x]);
}
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&c[i]);for(int i=1;i<=n;i++){int x,y;scanf("%d%d",&x,&y);addl(x,y);addl(y,x);}dfs(0,0);if(n<=1000&&m<=1000){while(m--){memset(v,0,sizeof(v));int x,y,a,b;scanf("%d%d%d%d",&x,&y,&a,&b);num=0;LCA(x,y);printf("%d\n",num-(v[a]&&v[b])); }return 0;}int t=(int)sqrt(cnt);for(int i=1;i<=m;i++){scanf("%d%d%d%d",&q[i].l,&q[i].r,&q[i].a,&q[i].b);q[i].l=rfn[q[i].l];q[i].r=rfn[q[i].r];if(q[i].r<q[i].l) swap(q[i].l,q[i].r);q[i].id=i;q[i].pos=(q[i].l-1)/t+1;}sort(q+1,q+1+m);int l=1,r=0;for(int i=1;i<=m;i++){while(r<q[i].r)	add(c[dfn[++r]]);while(l<q[i].l)	del(c[dfn[l++]]);while(l>q[i].l)	add(c[dfn[--l]]);while(r>q[i].r)	del(c[dfn[r--]]);ans[q[i].id]=num-(v[q[i].a]&&v[q[i].b]);}for(int i=1;i<=m;i++)printf("%d\n",ans[i]);
}

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$T210$分$code$

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[81],A,B,num[10],k,a[10];
int dfs(int x,int limst,int sum)
{if(!x){if(sum&1) return 0;int mid=sum/2;f[0]=1;for(int i=1;i<=k;i++)for(int j=mid;j>=a[i];j--)f[j]|=f[j-a[i]];if(f[mid]) return 1;return 0;}int r=limst?num[x]:9,ans=0;for(int i=0;i<=r;i++){a[x]=i;ans+=dfs(x-1,limst&&(i==r),sum+i);}return ans;
}
int solve(int x)
{k=0;while(x){num[++k]=x%10;x/=10;}return dfs(k,1,0);
}
int main()
{scanf("%d%d",&A,&B);printf("%d",solve(B)-solve(A-1));
}

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总结

$T1$开始以为是树链剖分然后发现不会剖，去看$T2$数位$dp$直接跳谢谢，然后$T3$以为是$dusontree$。开始写题，$T1$分块显然不行，然后十分显然的树上莫队。但是不会就瞎敲，知道要用欧拉序，但是写了半天自己钦定树上莫队不可做(脑子抽了)。然后去写了三题的暴力。

然后结果就四个字！$$\begin{gathered} \text{暴力写挂} \\ \text{垫底了呀} \end{gathered}$$

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$$\text{：<}$$

$$\begin{gathered} \text{纪你太美} \\ \text{jzoj you so beautiful} \end{gathered}$$