Hello, guys. Can you tell me, why i got TL for this?


#define _USE_MATH_DEFINES

#include <stdio.h>
#include <math.h>

struct Point
{
    float x;
    float y;
};

int main()
{
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);

    int nPins = 0;
    float pinRadius = 0;
    scanf("%d %f\n", &nPins, &pinRadius);

    Point points[101];
    for (int i = 0; i < nPins; i++) {
        scanf("%f %f\n", &points[i].x, &points[i].y);
    }

    float len = 2*M_PI*pinRadius;
    for (int i = 1; i < nPins; i++) {
        len += sqrt(pow(points[i].x - points[i-1].x, 2) +
                    pow(points[i].y - points[i-1].y, 2));
    }
    len += sqrt(pow(points[0].x - points[nPins-1].x, 2) +
                pow(points[0].y - points[nPins-1].y, 2));

    printf("%.2f\n", len);
    return 0;
}

do not forget about {$H+} after debuging ;^)

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
    const unsigned int N=32768;
    unsigned long long xl[N];
    unsigned int col=0;
    cin.setf(ios::skipws);
    while(cin>>xl[col]) col++;
    cout.unsetf(cout.flags());
    cout.setf(ios::showpoint|ios::fixed|ios::dec);
    while(col--) cout<<setprecision(4)<<sqrt((double)xl[col])<<endl;
    return 0;
}

my program works correctly on all the tests from the forum...
//I use long arithmetic and DP

Good morning!There is a task № 1601. http://acm.timus.ru/problem.aspx?space=1&num=1601
I can't understand how to limit data input?The condition is that we can see "line feed", but this "line feed" is the end of data input in the function "std::getline". Which of symbols must show whether there is the end of data input excepting "line feed"? There is no explanation in the task what must be this symbol.

Edited by author 28.06.2009 13:33

Edited by author 28.06.2009 13:33

I solved this problem with O ( N ^ 2 lg N) for 1.781 sec :)
Just I use finding the angle of every line with Ox, Oy, Oz.
Equation of line is WRONG - O ( N ^ 3).

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
const double eps=1e-9;
struct point{long x,y,z;};
struct f
{
double a,b,c;
bool operator==(f x)
{return fabs(x.a-a)<eps&&fabs(x.b-b)<eps&&fabs(x.c-c)<eps;}
};
double sqr(double t){return t*t;}
bool cmp(f a,f b)
{
if(a.a!=b.a)return a.a-b.a<eps;
else if(a.b!=b.b)return a.b-b.b<eps;
else return a.c-b.c<eps;
}
bool cmp1(point a,point b)
{
if(a.x!=b.x)return a.x-b.x<eps;
else if(a.y!=b.y)return a.y-b.y<eps;
else return a.z-b.z<eps;
}
int main()
{
int n,mxx=-1;
point a[2000];
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z);
if(n==1){cout<<1<<endl;return 0;}
sort(a,a+n,cmp1);
for(int i=0;i<n-1;i++)
{
f m[2000];
int k=0;
for(int j=i+1;j<n;j++)
{
double d=sqrt(sqr(a[j].x-a[i].x)+sqr(a[j].y-a[i].y)+sqr(a[j].z-a[i].z));
if(d>0.0){m[k].a=(a[j].x-a[i].x)/d;m[k].b=(a[j].y-a[i].y)/d;m[k].c=(a[j].z-a[i].z)/d;k++;}
}
sort(m,m+k,cmp);
int br1=1,mx=-1;
for(int j=1;j<k;j++)if(m[j]==m[j-1])br1++;else{mx=max(mx,br1);br1=1;}
mx=max(mx,br1);
mxx=max(mxx,1+mx);
}
printf("%d\n",mxx);
return 0;
}

find #5 test, Please

#include <iostream.h>
main()
{int a,i,b,x;
cin>>a;
for(i=a; ;i++) {x=i;b=0;while(x!=0) {b=b*10+x%10; x=x/10;}
if(b==i) break;} cout<<i; return 0;}

I receive AV(CRASH) all the time. So, is there are someone who solve this problem on C?

Best Regards,
Veniamin.

What will be the output for N=10?

in:
10 3 5
out:
2

Tell my please, why I have Wrong Answer on test number 13 and if it is not problem, tell how I can solve this problem to e-mail WitaliyProg@mail.ru
Thanks!

It's very easy.
TEST#1 is a data which result is -1(It means "no answer").
I submit my program ,and accept!

Edited by author 07.05.2009 17:59

Сабж. Условие не соответствует решению

I haven't crash on my computer on all tests but have crash floating-point 10... Also my answers r<20000 is right if my brute force is right. sorry for my bad English

I write program and get WA3:
the number must be minimal.

O(N) solution ;)

Попытался разобраться как строить треугольник A2B2C2 и случайно додумался вот до такого "доказательства" теоремы Бориса:
0. Дано:
    A1B1 = A2B2,
    B1C1 = B2C2,
    ∠ B1A1C1 = ∠ B2A2C2.
1. По теореме синусов из перечисленного в п. 0 находим, что sin(∠ B2С2A2) = (A2B2 * sin(∠ B2A2C2)) / B2C2 и sin(∠ B1С1A1) = (A1B1 * sin(∠ B1A1C1)) / B1C1. Значит угол ∠ B2С2A2 = ∠ B1С1A1.
2. Соответственно, оставшийся угол A1B1C1 = A2B2C2 (следствие из суммы углов в треугольнике).
3. Теорема доказана!

Если найдете ошибку в доказательстве - значит решите задачу ;)

What's so cool in this test? It pushes a lot into one stack or a lot into many stacks or what?
#include <iostream>
using namespace std;

struct elem
{
    unsigned int value;
    elem * next;
};
elem ** tree;
elem * push(elem * stack,unsigned int value)
{
    if(!stack)
    {
        stack = new elem;
        stack->value = value;
        stack->next = NULL;
        return stack;
    }
    else
    {
        elem * temp = new elem;
        temp->value = value;
        temp->next = stack;
        return temp;
    }
}
elem * pop(elem * stack,unsigned int & value)
{
    elem * temp = stack->next;
    value = stack->value;
    delete stack;
    return temp;
}
int main()
{
    tree = new elem* [1000];
    memset(tree,0,1000*4);
    int n;
    cin>>n;
    char str[5];
    int a;
    unsigned int b;
    for(int i = 0;i<n;++i)
    {
        cin>>str;
        if(str[1]=='U')
        {
            cin>>a;
            cin>>b;
            tree[a-1] = push(tree[a-1],b);

        }
        else
        {
            cin>>a;
            unsigned int value;
            tree[a-1] = pop(tree[a-1],value);
            cout<<value<<endl;

        }
    }

    return 0;
}