any one help??  I need help...

I'm sure my source code is perfect. But it never passed test #5. Can anyone help me?

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

int main()
{
        double  cx,cy,cz,nx,ny,nz,r,sx,sy,sz,vx,vy,vz,t;
        double  x,y,z;
        double  a,b,c;
        int     hit;
        scanf("%lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf %lf",&cx,&cy,&cz,&nx,&ny,&nz,&r,&sx,&sy,&sz,&vx,&vy,&vz);
        b=nx*vx+ny*vy+nz*vz;
        c=nx*(sx-cx)+ny*(sy-cy)+nz*(sz-cz);
        a=-5*nz;
        if(b*b-4*a*c<-1e-8)
                printf("MISSED\n");
        else if((b+sqrt(b*b-4*a*c))/nz<-1e-8 && (b-sqrt(b*b-4*a*c))/nz<-1e-8)
                printf("MISSED\n");
        else
        {
                hit=0;
                t=(b-sqrt(b*b-4*a*c))/(10*nz);
                if(t>1e-8)
                {
                        x=sx+vx*t;
                        y=sy+vy*t;
                        z=sz+vz*t-5*t*t;
                        if(r-sqrt(pow(x-cx,2)+pow(y-cy,2)+pow(z-cz,2))>1e-8)
                                ++hit;
                }

                t=(b+sqrt(b*b-4*a*c))/(10*nz);
                if(t>1e-8)
                {
                        x=sx+vx*t;
                        y=sy+vy*t;
                        z=sz+vz*t-5*t*t;
                        if(r-sqrt(pow(x-cx,2)+pow(y-cy,2)+pow(z-cz,2))>1e-8)
                                ++hit;
                }
                if(hit)
                        printf("HIT\n");
                else
                        printf("MISSED\n");
        }
        return 0;
}

I solved it in 0.015 with g++, but i can't optimize anymore

Edited by author 22.06.2015 13:32

It's just the array rotation problem asked of a different way.

Please help me with test case 16

Hey guys any hints here? All examples from forum works on my program but this one
exceeds. I guess it's because more than 200 characters are written, but I cannot find any test case that will require less written characters but the output will have more length

Does anyone know some tests to overcome WA 2?

I use a sement tree,  gave tle in test 17 and when I changed the scanf by the following "std :: ios :: sync_with_stdio (false)" I have acepted with time of 0.358 s

Since there can be more than one acceptable solution, I think, it could give a more unambiguous result.

who could give some testdata to me ?


3  3
9
10
11

result?

Edited by author 05.11.2008 04:11

using System;

class Program
{
    static void Main()
    {
        int t = int.Parse(Console.ReadLine());
        string[] s = Console.ReadLine().Split(' ');
        Array.Sort(s);
        int r = 0;

        for (int i = 0; i < (t / 2)+1; i++)
        {
            r += ((int.Parse(s[i]) / 2) +1);
        }
        Console.WriteLine(r);
    }
}

Edited by author 16.06.2012 19:10

Edited by author 16.06.2012 19:11

Hi, folks!

Can anyone give me the source code in C# of this task (mine is already AC) when the timing is less than 0.1 sec?
Or can anyone give me the key notes how it can be decreased in almost ten times?

Thanks in advance!

#include <iostream>
using namespace std;

int main() {
    int cand, ch;
    cin >> cand >> ch;
    int* Ch = new int[ch];
    double* Cand = new double[cand];

    for (int i = 0; i < cand; i++) {
        Cand[i] = 0;
    }

    for (int i = 0; i < ch; i++)
        cin >> Ch[i];

    for (int i = 0; i < ch; i++) {
            Cand[Ch[i]-1] += 1;
    }

    for (int i = 0; i < cand; i++) {
        printf("%.2f%%\n", (Cand[i]/6.0)*100.0);
    }
}
+++++++++++
Why this isn't work?

#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;

int main() {
    int i,j,N,p;
    std::cin>>N;

    float cd[N][2],R,dist = 0.0;
    std::cin>>R;

    for(p = 0;p < N;p++)
        std::cin>>cd[p][0]>>cd[p][1];

    i = 0;
    j = i + 1;

        dist = (2 * std::acos(-1.0) * R);

    while(i < N)
    {
        dist += std::sqrt( ( (cd[j][0] - cd[i][0]) * (cd[j][0] - cd[i][0]) ) +
( (cd[j][1] - cd[i][1]) * (cd[j][1] - cd[i][1]) ) );
        i++;
        j = (i + 1) % N;
    }

    std::cout <<std::setprecision(2) << std::fixed <<dist;
    return 0;
}

What answer should be for given test?

1 1 1 1
0

Can anyone explain the sample:

3
0 0 0
1 1 0
1 1 0



1
1 3


I dont understand it. The resulting matrix is:
0 1 1
0 1 1
0 0 0

Thanks!

Same solution:

Visual C++ 2013:
Accepted    0.686    8 120 КБ

G++ 4.9 C++11:
Time limit exceeded    21    1.513    3 924 КБ

oh my code..

first  use dynamic programming to find the minimum value of distance from 0 and visit 1~m once then to 0, we can get a chain 0--i1-i2--...-im--0

then we consider to cut this chain, use dynameic programming dp[i] to record distance.we cut first i node of the chain..
dp[m] is the result..

I'll try this idea...

rt

can someone explain test number 2?