j-1 bit must be 1 if prefix of the string could be cut by j palindroms. in notation 9-->1110 , but 9 must be 10000000, isn`t it???

I was using QUEUE.
It was Memory limit exceed #8 or something similar.
Replaced queue by simple an static array of one million elements.
I was using state struct.
There are three members of unsigned char type.
So,  with C limitations, I got AC with 16 MB.

На руби вообще не имеет смысла что-то решать?

add test x = y; x + y < c; x, y < c
example input: 900000000 900000000 1000000000, i had AC with output 900000000 900000000

When I was using STD::SET and STD::MAP,  it was TLE #18.
When I switched to STD::UNORDERED_SET and STD::UNORDERED_MAP,  it was ACCEPTED 0.6 sec 38 MB.
I was using reserve (1<<16) and max load factor of 0.25.

Hello,
Does anybody know what is the 2nd test? I can't get my head round this issue.

If you want to run your own contest system, you can try my open-source server: http://codejudge.azurewebsites.net
Source code: https://github.com/Backs/judge.net
Work in progress, I'll add new features.
If you want to try it, you can write me, I'll help to install and configure it.

Братаны-кодеры в этой задаче нужен особый подход к считыванию
Из условия неясно,какой формат,но пока что он следущий(.=пробел):
ххх.ххх.ххх и т.д.
ууу.ууу.ууу.ууу и т.д.
Не используйте на Паскале eof,eoln.
Часть моего говнокода:
 ...
 s:string[4];
 ...
 len:=1;
 read(s);
 inc(c[90*90*(ord(s[1])-33)+90*(ord(s[2])-33)+(ord(s[3])-33)]);
 f:=length(s)=4;
 while f do begin
  inc(len);
  read(s);
  inc(c[90*90*(ord(s[1])-33)+90*(ord(s[2])-33)+(ord(s[3])-33)]);
  f:=length(s)=4;
 end;
 ...

C++.  I was using recursive DFS. vector <vector<int> >.   I forgot to pass graph vector by reference and got MLE#8. After noticing that I am passing graph vector by value and simply adding an ampersand before vector name AC 1MB memory used.

Edited by author 03.07.2017 15:43

98000 3334 3334
1
25000 3333
3
1 97999 1
2 30000 1
2 25000 1

Answer - You should rent the apartment #1 alone. or You should rent the apartment #2 alone. or You should rent the apartment #3 alone.

21111 1111 7777
5
1111 7777
10000 6556
20000 3131
80000 2
76400 1
4
1 21110 6000
1 21111 6665
2 21111 1220
2 1110 0

Answer - You should rent the apartment #3 alone.

1001 1001 1001
1
1001 1
1
2 1001 1

Answer - You should rent the apartment #1 alone.

I hope it can help. My program, which got WA 6, failed on this tests.

My AC program gives the wrong answer on the test:
2 1A 1A
1
5 1B 1A 1A 1B 1B
=>0.200000000

but rught answer 0.400000000

I was trying C++ for several hours.  Can't defeat test case No.  7. So,  I gave up and got AC using Python Decimal module just after several minutes I stopped trying C++.  It is bad I think that I don't solved with C++.  Literally no idea why WA#7. I tried rounding and truncation.  The result was always the same.

When time is 02:00:00 Good have power=58 and Evil=60 ((x-40)/(130-40)=7200/36000 => x=58).

In your segment tree you have count of zeros on every segment of vector of length N.
Find k-th non-zero element.  Then zero it.  Update tree.  Output.

Sort by T1.
Then set timer=0,  ans =0.
Then for all students in sorted order:
if timer<T1 then timer=T1
timer=timer+T2
dtime=timer-T3
if dtime>ans then ans=dtime

https://technology.cpm.org/general/3dgraph/
Just in case someone (like me) needs a handy tool for 3D points visualization

Don't know, what's wrong. Give me tests please.

a=[]
n=int(input())
for i in range(0,n):
    f=(int,input())
    a.append(f)
sum=0
ans=0
for i in range(0,n-2):
    sum=0
    for f in range(i,i+3):
        sum=sum+a[f]
    if(sum>ans):
       ans=sum
       z=f
print(ans)
print(z)


#What could be the error in this. It works on my computer. Why do they show runtime error #when submitted??

long long at least, otherwise WA12. doesn't fit even in unsigned long

What's wrong?
#include<cstdio>
using namespace std;
const int N=201;
struct Node{int x,y;}a[N];
struct Line{int s,t;}b[N];
int fa[N],n,m,root;
inline int direct(Node x,Node y,Node z){
    return (y.x-x.x)*(z.y-x.y)-(y.y-x.y)*(z.x-x.x);
}
bool online(Node x,Line y){//right
    Node be=a[y.s],en=a[y.t];
    if (x.x<be.x||x.x>en.x) return 0;
    if (direct(be,x,en)==0) return 1;
    return 0;
}
bool cross(Line x,Line y){
    Node p1=a[x.s],p2=a[x.t],p3=a[y.s],p4=a[y.t];
    if (online(p1,y)||online(p2,y)||online(p3,x)||online(p4,x)) return 1;
    int d1=direct(p1,p2,p3),d2=direct(p1,p2,p4),d3=direct(p3,p4,p1),d4=direct(p3,p4,p2);
    if ((d1^d2)<0&&(d3^d4)<0) return 1;
    return 0;
}
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void unite(int x,int y){
    x=find(x); y=find(y);
    if (x<y) fa[y]=x; else if (x>y) fa[x]=y;
}
int main(){
    scanf("%d%d",&n,&m);
    for (int i=1; i<=n; i++) fa[i]=i;
    for (int i=1; i<=n; i++) scanf("%d%d",&a[i].x,&a[i].y);
    for (int i=1; i<=m; i++){
        scanf("%d%d",&b[i].s,&b[i].t); unite(b[i].s,b[i].t);
        for (int j=1; j<=n; j++)if (online(a[j],b[i])) unite(j,b[i].s);
        for (int j=1; j<i; j++) if (cross(b[i],b[j])) unite(b[i].s,b[j].s);
    }
    root=find(1);
    for (int i=2; i<=n; i++) if (find(i)!=root) {printf("NO"); return 0;}
    printf("YES");
    return 0;
}