Xot kto nibud, skajite, chto takoe sluchaet test13

It is possible to get AC with a 10-lines program. Simply use regular expressions:

Q = "(?<!\\\\)\"" for " which is not preceded by \

".*?(?:\n\r?\n|\\\\par(?![A-Za-z])|$)" for paragraph (note \r)

For each paragraph you get with Scanner.findWithinHorizon, do replaceAll matching Q+"(.*?)"+Q with "``$1''" and then replaceFirst Q with an empty string.

Btw, the real TeX parsing is much more complicated, e.g., \\par would not end a paragraph.

I passed test 3 after replacing my "blank line" regexp "\n\n" with "\n\r?\n". Apparently "blank" (which used to mean "empty") here means empty or "\r".

please give some tests, WA7

I've check all the data here..but i still get WA#12..

#include <iostream>
#include <cstring>
#include <string>
#define inf 214748364
using namespace std;
struct tmono
{
    int n[9000];
    int w,nega,xs;
};
char s[1000];
int xs,len,con,e;
tmono num;
string res;
inline bool isnum(char ch)
{
    return '0'<=ch&&ch<='9';
}
inline bool getnum(int st,int en,tmono &e)
{
    memset(e.n,0,sizeof(e.n));
    e.w=0;e.xs=0;e.nega=0;
    if(s[st]=='-')    e.nega=1,++st;
    else
    if(s[st]=='+')    e.nega=0,++st;
    /*
    while(s[st]=='+'||s[st]=='-')
    {
        if(s[st]=='-')
            e.nega=!e.nega;
        ++st;
    }
    */
    for(int i=en;i>=st;i--)
    if(s[i]=='.')
    {
        e.xs=en-i;
        for(int j=i;j<en;j++)
            s[j]=s[j+1];
        --en;
        break;
    }
    for(int i=st;i<=en;i++)
    if(!isnum(s[i]))
        return 0;
    e.w=0;
    for(int i=en;i>=st;i--)
        e.n[++e.w]=s[i]-'0';
    return 1;
}
inline bool work()
{
    res="";
    cin>>xs;
    int tempxs=xs;
    scanf("\n");
    if(s[0]=='.')
    {
        for(int i=len-1;i>=0;i--)
            s[i+1]=s[i];
        s[0]='0';
        len++;
    }
    if(!isnum(s[0])&&s[0]!='-'&&s[0]!='+')
        return 0;
    con=0;
    for(int i=0;i<len;i++)
    {
    if(s[i]=='e'||s[i]=='E')
    {
        if(con!=0)
            return 0;
        con=i;
    }
    if(s[i]!='e'&&s[i]!='E'&&s[i]!='-'&&s[i]!='+'
        &&s[i]!='.'&&!isnum(s[i]))
            return 0;
    if(s[i]=='.'&&!isnum(s[i+1]))
        return 0;
    }
    e=0;
    bool enega=0;
    if(con!=0)
    {
        e=-inf;
        int j=con;
        con++;
        if(s[con]=='+'||s[con]=='-')
        {
            if(s[con]=='-')    enega=!enega;
            con++;
        }
        /*
        while(s[con]=='-'||s[con]=='+')
        {
            if(s[con]=='-')    enega=!enega;
            ++con;
        }*/
        for(int i=con;i<len;i++)
        {
            if(!isnum(s[i]))
                return 0;
            if(e==-inf)
                e=0;
            e=e*10+s[i]-'0';
            if(e>500)
                break;
        }
        if(e==-inf)
            return 0;
        if(enega)
            e*=-1;
        len=j;
    }
    --len;
    if(!getnum(0,len,num))
        return 0;
    num.xs-=e;
    while(num.n[num.w]==0&&num.w>0)
        --num.w;
    bool isxs=0;
    if(num.xs<=0)
    {
        for(int i=num.w;i>=1;i--)
            res+=num.n[i]+'0';
        while(num.xs<0)
        {
            res+="0";
            num.xs++;
        }
    }else
    {
        int j=max(num.w,num.xs);
        if(num.xs>num.w)
        {
            num.w=num.xs;
            num.xs=0;
            res+="0.";
            isxs=1;
        }
        for(int i=j;i>=1;i--)
        {
            if(i==num.xs)
                isxs=1;
            if(isxs&&xs==0)
                break;
            if(i==num.xs)
                res+=".";
            if(isxs)--xs;
            res+=num.n[i]+'0';
        }
    }
    while(xs>0)
    {
        if(!isxs)
            res+=".",isxs=1;
        res+="0",--xs;
    }
    while(res[0]=='0')
        res.erase(0,1);
    if(res.length()==0)
        res.insert(0,"0");
    if(res[0]=='.')
        res.insert(0,"0");
    for(int i=0;i<res.length();i++)
    if(res[i]=='.')
        if(res.length()-i<tempxs)
        {
            tempxs-=res.length()-i;
            while(tempxs--)
                res.insert(res.length()-1,"0");
        }else
            res.erase(i+tempxs,res.length()-i-tempxs-1);
    if(res[res.length()-1]=='.')
        res.erase(res.length()-1,1);

    if(num.nega)
    for(int i=0;i<res.length();i++)
    if('1'<=res[i]&&res[i]<='9')
    {
        res.insert(0,"-");
        break;
    }
    for(int i=1;i<res.length();i++)
    if(res[i]=='.'&&!isnum(res[i-1]))
    {
        res.insert(i,"0");
        break;
    }
    if(res[res.length()-1]=='.')    res.erase(res.length()-1,1);

    res+="\n";
    return 1;
}
inline void reads()
{
        memset(s,0,sizeof(s));
        gets(s);
        len=strlen(s);
        while(len>=1&&isspace(s[len-1]))
            s[len]=0,--len;
        while(isspace(s[0]))
        {
            for(int j=0;j<len-1;j++)
                s[j]=s[j+1];
            s[len]=0;len--;
        }
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("input.in","r",stdin);
    freopen("output.out","w",stdout);
    #endif
    reads();
    while(!(strlen(s)==1&&s[0]=='#'))
    {
        if(!work())
            printf("Not a floating point number\n");
            else
            cout<<res;
        reads();
    }
    return 0;
}

NOT FLOATING FOR
1)1E2.1

2)2E2+1

3)2E2-1

4)E

5)-+3

6)3EE

-----------------------------------------
1)
1E1 1
10.0

1E+1 1
10.0

0E10 1
0.0

0 1
0.0

000.1 1
0.1

0.1E1 0
1

Is my greedy algo correct?
On each step I choose an edge to vertex with the greatest out-degree.
I have WA 8.

I know AC solution that counts number of cw and ccw moves from point i to i+1 seen from the first point. Here is the test on which this AC solution answers cw but correct answer is ccw.

17
0 0
10 0
10 10
11 10
11 9
11 8
11 7
11 6
11 5
11 4
11 3
11 2
11 1
11 0
12 0
12 11
0 11

I don't understand this problem. As I think, the third sample is the tree from the picture. But I can not understand why, for example, 'the number of the ancestor of this node' for the node #18 is 12? What does it mean, 'the number of the ancestor of this node'?

THank you.

small a,
big b,
big n,
and the LCM of a and b is larger than n.
such as
3 24999998 25000000


Edited by author 21.03.2010 23:41

I have written 2 programs and both get WA19
It is very strange, I am already tired to write programs to this problem.

first      2*BFS
secondar   Dijkvasta+Heap
both got Wa19 i not understand why.
Somebody help me?

there is my program


var
c:array[0..1800] of real;
n,m,i ,fn:longint;
begin
readln(n,m);
c[0]:=1;
c[1]:=m-1;
i:=1;
repeat
i:=i+1;
{for i:=2 to n do   }
c[i]:=(m-1)*(c[i-1]+c[i-2]);
until i>=n;
writeln(c[n]:0:0);
end.

My AC program runs about 10 seconds on following tests;

4 41
4 49 (on this test it works more, that 1 minute!!!)
17 43
19 49
41 43
50 41
50 43

I can write about 10 more tests, but, I think, it's enough:)
Could you add some of these tests?

Here is my code, what can make my code run faster? Please check it up...

#include <iostream>
#include <cmath>
using namespace std;
bool prime(long long x)
{
    int i;

    if(x==2)
       return true;
    else if(x%2==0)
       return false;

    for(i=3; i*i<=x; i+=2)
        if(x%i==0)
           return false;
    return true;

}
int main()
{
    long long K, s;

    cin>>K;
    if(prime(K)==true)
    {
        cout<<K-1;
        return 0;
    }
    else
    {
        for(long long i=3; i<499999950; i++)
        {
            if(K%i==0)
            {
                cout<<i-1;
                return 0;
            }
        }
    }
    return 0;
}

Hope this test can help you:

2
3
2 1 0
1
0
-
NO
YES

Good luck!

why my program doesnot work on WA#2????
var a:array [1..10000] of integer;
i,k,s:integer;
begin
k:=0;
s:=1;
for i:=1 to 10 do
readln(a[i]);
for i:=1 to 10 do
s:=s*a[i];
for i:=1 to s do
if s mod i=0 then
k:=k+1;
if (k>0)and (k<10) then writeln(k);
end.

Edited by author 21.03.2010 05:16

I can't find a mistake in my solution. Can anybody give me any hint?

#include <cstdlib>
#include <iostream>
#include <iomanip>

using namespace std;

int main(int argc, char *argv[])
{

    double a,b,c,d,m,n=2,mn;
    cin>>a>>b;
    c=a/100;
    d=b/100;
    while (1)
    {m=1;
     while (m<n)
    {mn=m/n;
     if ((mn<d)&&(mn>c)) break;
     m++;}
     if (m!=n) break;
     n++;}
     cout<<n<<endl;
    system("PAUSE");
    return EXIT_SUCCESS;
}




This is my project Got WA on test14.
I think there's something wrong with digits
But i cannot find it
Can anyone help me with it?
Thx!

  I used "dp"  and spend 261k space and 0.046 time,

  but i see the rank 1 only used 0.015 second?

            How did he make it?