Who knows how to solve it?
:?)

#include<stdio.h>
void main(){
    int n,k=0,j;
    int result=1;
    char test[20];
    scanf("%d %s",&n,test);
    for (j=0;j<20;j++) if (test[j]=='!') k++;
    if(n%k==0){
        for(j=0;n-j*k!=k;j++){
            result*=(n-j*k);
        };
        result*=k;
    }else{
        for(j=0;n-j*k!=n%k;j++){
            result*=(n-j*k);
        };
        result*=n%k;
    };
    printf("%d",result);
}


this is my source code please tell me why my program is time limited exceeded

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <utility>

using namespace std;

const int maxN = 304;

int x[maxN], y[maxN];
int n;
vector< pair<int, int> > p;
bool mark[maxN][maxN];

int dist(const int& i, const int& j)
{
    return (x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]);
}

bool compairation(const pair<int, int>& p1, const pair<int, int>& p2)
{
    return (dist(p1.first, p1.second) > dist(p2.first, p2.second));
}

int ccw(const int& i, const int& j, const int& k)
{
    int dx1 = x[j] - x[i], dy1 = y[j] - y[i];
    int dx2 = x[k] - x[j], dy2 = y[k] - y[j];
    return (dx1 * dy2 - dx2 * dy1);
}

int main()
{
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> x[i] >> y[i];

    for (int i = 1; i < n; i++)
        for (int j = i + 1; j <= n; j++)
            p.push_back(pair<int, int>(i, j));

    sort(p.begin(), p.end(), compairation);

    memset(mark, true, sizeof(mark));

    double total = 0.0;
    for (int i = 0; i < p.size(); i++)
        if (mark[p[i].first][p[i].second])
        {
            int d = p[i].first, c = p[i].second;
            vector<int> point;
            point.clear();
            for (int j = 1; j <= n; j++)
                if (ccw(d, j, c) == 0) point.push_back(j);
            for (int u = 0; u < point.size(); u++)
                for (int v = u + 1; v < point.size(); v++)
                {
                    mark[point[u]][point[v]] = false;
                    mark[point[v]][point[u]] = false;
                    //cout << point[u] << " " << point[v] << endl;
                }
            total += sqrt(double(dist(d, c)));
        }

    cout << round(total) << endl;

    return 0;
}

<code deleted>

Edited by author 19.02.2008 00:29

  Please consider the Sample input and output :
  The 1st flight ( connects 1 and 2 ) and the 2-nd flight ( 2-3)
are flights that depart from one airport ( airport 2 ) ( The flight
are 2 directions flight ) . But the sample output indicates
that their numbers are 4 and 2 , so their greatest common divisor
can't be 1 as the problem description !
   I don't know if I wrong . I got WA too . Please explain for me
Thanks a lot ! .

this is my program :


#include<iostream>
using namespace std;

int main()
{
    int n;
    cin>>n;
    int i;
    for(i=0;i<n/2;i++)
        cout<<i+1<<" "<<n-i<<" ";
    if(n%2!=0)
        cout<<((int)n/2)+1<<" ";
    cout<<endl;
    cout<<"1 ";
    for(i=3;i<n;i+=2)
        cout<<i<<" ";
    int p=n;
    if(p%2!=0)
    {    p-=1;
        cout<<n<<" "; }
    for(p=p;p>1;p-=2)
        cout<<p<<" ";
    return 0;
}

I think its right but I get WA on test 4 anybody knows whats wrong ?

I have wr . Please give me some test
help help!!!!!

Don't brute force thats the trick ! :D

can you please help me about what is wrong about this?



You tried to solve problem 1161 Stripies.
Your solution was compiled with the following errors:

93e006e3-949c-4a74-a88f-eb4b2c6dc082
93e006e3-949c-4a74-a88f-eb4b2c6dc082(19): error: expression must have a constant value
int set[N];
^

/* why can't I declare a new array like this? (N is integer) */

93e006e3-949c-4a74-a88f-eb4b2c6dc082(61): error: expression must have a constant value
int aux[max-min+1], i=min, k=0;
^

/* The same */

93e006e3-949c-4a74-a88f-eb4b2c6dc082(61): error: expression must have a constant value
int aux[max-min+1], i=min, k=0;
^

/* Two errors on the same line!!! */

compilation aborted for 93e006e3-949c-4a74-a88f-eb4b2c6dc082 (code 2)

Here are some tests:


10
1
10 + 0 = 10

21
0

43
0

100
3
91 + 9 = 100
95 + 5 = 100
100 + 00 = 100

4444
6
3722 + 722 = 4444
4022 + 422 = 4444
4037 + 407 = 4444
4040 + 404 = 4444
4042 + 402 = 4444
4222 + 222 = 4444

6667
1
6061 + 606 = 6667

6668
6
5834 + 834 = 6668
6034 + 634 = 6668
6059 + 609 = 6668
6062 + 606 = 6668
6064 + 604 = 6668
6334 + 334 = 6668

66667
1
60607 + 6060 = 66667

66668
7
58334 + 8334 = 66668
60334 + 6334 = 66668
60584 + 6084 = 66668
60604 + 6064 = 66668
60608 + 6060 = 66668
60634 + 6034 = 66668
63334 + 3334 = 66668

52525252
11
46262626 + 6262626 = 52525252
47747626 + 4777626 = 52525252
47750126 + 4775126 = 52525252
47750176 + 4775076 = 52525252
47750226 + 4775026 = 52525252
47750231 + 4775021 = 52525252
47752626 + 4772626 = 52525252
47762626 + 4762626 = 52525252
47812626 + 4712626 = 52525252
48262626 + 4262626 = 52525252
51262626 + 1262626 = 52525252

456000
10
378000 + 78000 = 456000
408000 + 48000 = 456000
413000 + 43000 = 456000
414500 + 41500 = 456000
414545 + 41455 = 456000
414546 + 41454 = 456000
414550 + 41450 = 456000
414600 + 41400 = 456000
415000 + 41000 = 456000
428000 + 28000 = 456000

123000321
1
111818474 + 11181847 = 123000321

200020002
16
150010001 + 50010001 = 200020002
181510001 + 18510001 = 200020002
181835001 + 18185001 = 200020002
181836351 + 18183651 = 200020002
181836366 + 18183636 = 200020002
181836371 + 18183631 = 200020002
181836401 + 18183601 = 200020002
181836501 + 18183501 = 200020002
181837001 + 18183001 = 200020002
181840001 + 18180001 = 200020002
181860001 + 18160001 = 200020002
181910001 + 18110001 = 200020002
182010001 + 18010001 = 200020002
185010001 + 15010001 = 200020002
190010001 + 10010001 = 200020002
200010001 + 00010001 = 200020002

1000000000
13
905000000 + 95000000 = 1000000000
909050000 + 90950000 = 1000000000
909090500 + 90909500 = 1000000000
909090905 + 90909095 = 1000000000
909090910 + 90909090 = 1000000000
909090950 + 90909050 = 1000000000
909091000 + 90909000 = 1000000000
909095000 + 90905000 = 1000000000
909100000 + 90900000 = 1000000000
909500000 + 90500000 = 1000000000
910000000 + 90000000 = 1000000000
950000000 + 50000000 = 1000000000
1000000000 + 000000000 = 1000000000

Should i use long ariphmetics???
How should i correctly calculate the value of polynom for given variable?

I solved this problem, but my program is wrong at 4 test.
Where is mistake?

var
l,k,j,m,n,mistake,o,sum,p,w:integer;  text1:string;
z,text:array[1..100]of string;
begin  o:=1;
repeat
inc(l);
readln(z[l]);
until z[l]='#';
dec(l);
repeat
inc(j);
readln(text[j]);
until text[j]=''; dec(j);
for k:=1 to l do
for m:=1 to j do
 begin
for n:=1 to length(text[m]) do
begin      text1:='';
while text[m][o] in ['a'..'z'] do
begin
text1:=text1+text[m][o]; inc(o);
end;
if length(text1)=length(z[k]) then
 begin
for w:=1 to length(z[k]) do
if text1[w]<>z[k][w] then inc(mistake);
if mistake=1 then
 begin
 inc(sum);
 for p:=1 to length(z[k]) do
  begin
  text[m][o-length(text1)+p-1]:=z[k][p];
  end;
 end;
mistake:=0;
end;  inc(o);
end;
o:=1;
(*----------------------*)
end;
for k:=1 to j do
writeln(text[k]);
writeln(sum);
end.

My program doesn't work on this test. Please, tell me what's wrong with it:

#include <stdio.h>

int C[500][500],R[500][500],U[500],T[500];

int main(int argc, char* argv[])
{
    int N,M,start,finish,s,a,b,c,S,F,i,j,u;

    scanf("%d%d",&N,&M);
    start=N;
    s=N-1;

    for (i=0;i<M;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        C[a-1][b-1]=c;
        T[a-1]++;
    }

    scanf("%d%d",&S,&F);
    S--;
    F--;
    U[s]=F;

    for (i=0;i<N;i++)
        if (i!=S&&i!=F&&T[i]==0)
        {
            s--;
            U[s]=i;
        }

    do
    {
        finish=start-1;
        start=s;

        for (i=finish;i>=start;i--)
        {
            b=U[i];
            for (j=0;j<N;j++)
                if (C[j][b]!=0&&j!=S)
                {
                    T[j]--;
                    if (T[j]==0)
                    {
                        s--;
                        U[s]=j;
                    }
                }
        }
    }
    while (s<start);

    U[0]=S;

    for (i=0;i<N;i++)
    {
        a=U[i];
        for (j=0;j<N;j++)
            R[i][j]=C[a][j];
    }

    for (i=0;i<N;i++)
    {
        b=U[i];
        U[i]=-1;
        for (j=0;j<N;j++)
            C[j][i]=R[j][b];
    }

    U[N-1]=0;

    for (i=N-2;i>=0;i--)
    {
        for (j=i+1;j<N;j++)
            if (C[i][j]!=0&&U[j]!=-1)
            {
                u=C[i][j]+U[j];
                if (u>U[i])
                    U[i]=u;
            }
    }

    if (U[0]==-1)
        printf("No solution");
    else
        printf("%d",U[0]);

    return 0;
}

please give me some tests!

#input
0.0 0.0 2.0 0.0
0
#output
3
0.0 0.0
1.0 0.0
2.0 0.0

This output is correct, isn't it?

Corrected. Should be %I64d for long long variables, not %lld. Now I have AC.
-----------------------------------------------------------
  printf("%lld %lld",max_wynik,min_wynik);  <---- BAD
  printf("%I64d %I64d,max_wynik,min_wynik); <---- GOOD

Edited by author 15.02.2008 02:51

What's wrong?

#include <stdio.h>
#include <stdlib.h>
int n,x,s;
typedef struct{
        int t1,t2,t3;
}st;
st t[110];
int max(int a,int b){
if(a>b)return a;
else return b;
}
int cmp(const void *a,const void *b){
        const st *c=(st *)a, *d=(st *)b;
        return c->t1 - d->t1;
}
int main(){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
                scanf("%d %d %d",&t[i].t1,&t[i].t2,&t[i].t3);
        }
        qsort(t,n,sizeof(st),cmp);
        for(int i=0;i<n;i++){
                t[i].t1=max(t[i].t1,x);
                x=t[i].t1 + t[i].t2;
                s+=max(0,x-t[i].t3);
        }
        printf("%d\n",s);
return 0;
}

Some limitations are identical, and when I used counter for entering edges to each vertex, I get WA15).


Edited by author 28.01.2008 12:26

what's the answer of 5  =  ="  Can u tell me..?

I have no hint.. T_T

Can i accept by use relation of sequence?

I tested my program several times but I can't find what's wrong with it please help !
my program :


#include<iostream>
#include<stdio.h>
using namespace std;

int main()
{
    char a[1000000];
    cin>>a;
    int i;
    int counter=0;
    int maxint=0;
    for(i=0;i<strlen(a);i++)
    {
        if(a[i]>64){
            if(maxint<a[i]-55)
                maxint=a[i]-55;
            counter+=a[i]-55; }
        else {
        if(maxint<a[i]-48)
                maxint=a[i]-48;
            counter+=a[i]-48; }
    }
    for(i=1;i<36;i++)
    {
        if((counter%i==0)&&(maxint<i))
        {
            cout<<i+1;
            return 0;
        }
    }
    cout<<"No solution.";
    return 0;

}