CodeVS第二次月赛

传送门

http://codevs.cn/problem/4109/
http://codevs.cn/problem/4110/
http://codevs.cn/problem/4111/

T1 TTY loves numbers

题目大意

给定一个模数m，构造一个最小01串，使其mod m=0
长度为len的01串满足，从第1位到第i位都为1，第i+1到第len位都为0 (1<=i<=len)

题解

一个11111...111的串的话可以看成10len+10len−1+...101等比数列求和公式可以写成

10len−19

根据欧拉定理，10ϕ(m)≡1(mod  m)
要求10与m互质，那么我们假设m为1e7以内的最大素数10999997
所以长度不超过1e7，一定有解不存在输出’Impossible’的情况
所以，我们从小到大构造全是1的串，如果能整除就直接输出，
如果不能找到之前与其取模之后相等的数做差即为答案

var x:array[0..11000005]of longint; i,j,k:longint; m,ans,len:longint;begin readln(m); ans:=1; len:=1; for i:=1 to 100000000 do  begin   if ans mod m=0   then begin     for j:=1 to len do write(1);     writeln; halt;   end   else    if x[ans mod m]<>0    then begin     for j:=1 to len-x[ans mod m] do      write(1);     for j:=1 to x[ans mod m] do      write(0);     writeln;     halt;    end    else x[ans mod m]:=len;   inc(len); ans:=(ans*10+1) mod m;  end;end.

T2 TTY Plants Sunflowers

题目大意

给你一棵n个点n-1条边的无根树和每个点的权值,询问任意两点间路线的LIS值

题解

回顾O(nlogn)的LIS,每次修改的是ans以及维护的长度为i的LIS结尾的最小值,dfs动态维护数组和ans就好了,离线处理处任意两点的答案
O(n2logn+m)

var w:array[0..8000,1..2]of longint; ans:array[0..2005,0..2005]of longint; d,x,y:array[0..2005]of longint; i,j,k:longint; n,m,len,a,b,anss:longint;procedure init(a,b:longint);begin w[len,1]:=b; if w[a,2]=0 then w[a,2]:=len else w[w[a,1],2]:=len; w[a,1]:=len; inc(len);end;function find(a:longint):longint;var l,r,mid:longint;begin l:=1; r:=anss; while l<r do  begin   mid:=(l+r)>>1;   if a<d[mid]   then r:=mid   else l:=mid+1;  end; exit(l);end;procedure dfs(a,b:longint);var tt,temp,e,f,g:longint;begin y[a]:=1; if x[a]>d[anss] then begin inc(anss); d[anss]:=x[a]; e:=anss; f:=0; g:=-1; end else begin temp:=find(x[a]); e:=temp; f:=d[e]; g:=0; d[e]:=x[a]; end; ans[b,a]:=anss; tt:=w[a,2]; while tt<>0 do  begin   if y[w[tt,1]]=0 then begin    dfs(w[tt,1],b);   end;   tt:=w[tt,2];  end; d[e]:=f; inc(anss,g);end;begin readln(n); len:=n+1; for i:=1 to n-1 do  begin readln(a,b); init(a,b); init(b,a); end; for i:=1 to n do  read(x[i]); for i:=1 to n do begin  fillchar(y,sizeof(y),0);  fillchar(d,sizeof(d),0);  anss:=0;  dfs(i,i); end; readln(m); for i:=1 to m do  begin   readln(a,b);   writeln(ans[a,b]);  end;end.

T3 TTY Saves Da’shgua

题目大意

给出一个序列，进行如下3个操作
1.对于第I个数a[i]变为x
2.对[L,R]区间内的每一个a[i],变为⌊ln(a[i])⌋
3.询问[L,R]内的和

题解

[BZOJ3211] 花神游历各国/[BZOJ3038] 上帝造题的七分钟2
⌊ln(a[i])⌋操作最多4次就变为0了，ln(0)=0，没有意义,所以我们对于操作2当某区间和为0时，不继续向下修改

var x:array[0..100000]of longint; w:array[0..600000,1..3]of int64; i,j,k:longint; n,m:longint; a,b,c:longint;procedure build(a,l,r:longint);var mid:longint;begin w[a,1]:=l; w[a,2]:=r;  if l=r  then begin w[a,3]:=x[l]; exit; end; mid:=(l+r)>>1; build(a*2,l,mid); build(a*2+1,mid+1,r); w[a,3]:=w[a*2,3]+w[a*2+1,3];end;procedure update1(a,b,c:longint);var mid:longint;begin if (w[a,1]=w[a,2])//and(w[a,1]=b) then begin w[a,3]:=c; exit; end; mid:=(w[a,1]+w[a,2])>>1; if b<=mid then update1(a*2,b,c) else update1(a*2+1,b,c); w[a,3]:=w[a*2,3]+w[a*2+1,3];end;procedure update2(a,l,r:longint);var mid:longint;begin if w[a,3]=0 then exit; if (w[a,1]=w[a,2]) then begin w[a,3]:=trunc(ln(w[a,3])); exit; end; mid:=(w[a,1]+w[a,2])>>1; if r<=mid then update2(a*2,l,r) else if l>mid then update2(a*2+1,l,r)  else begin update2(a*2,l,mid); update2(a*2+1,mid+1,r); end; w[a,3]:=w[a*2,3]+w[a*2+1,3];end;function query(a,l,r:longint):int64;var mid:longint;begin if (w[a,1]=l)and(w[a,2]=r) then exit(w[a,3]); mid:=(w[a,1]+w[a,2])>>1; if r<=mid then exit(query(a*2,l,r)) else if l>mid then exit(query(a*2+1,l,r))  else exit(query(a*2,l,mid)+query(a*2+1,mid+1,r));end;begin readln(n,m); for i:=1 to n do   read(x[i]); build(1,1,n); for i:=1 to m do  begin   readln(a,b,c);   case a of   1:update1(1,b,c);   2:update2(1,b,c);   3:writeln(query(1,b,c));   end;  end;end.