I dont understand: when I had swaped 'Impossible' to '0' my program solved more tests than late.

void fallBack(int angle, int M, int curEnergy, int curAmmo);
void advance(int angle, int M, int curEnergy, int curAmmo);
int findX(int curEnergy);

int main()
{
    int curEnergy, curAmmo;
    char behaviour;
    int M, MP, angle;
    int N, NP, offset;
    scanf_s("%i %i", &curEnergy, &curAmmo);
    scanf_s(" %c", &behaviour);
    scanf_s("%i %i %i", &M, &MP, &angle);
    if (behaviour == 'A') {
        scanf_s("%i %i", &N, &NP);
    }
    else if (behaviour == 'P') {
        scanf_s("%i", &offset);
        offset *= -1;

    }
    while (1) {
        switch (behaviour) {
        case 'A':
            if ((N*NP) > (M*MP * 3)) {
                advance(angle, M, curEnergy, curAmmo);
                return 0;
            }
            else {
                fallBack(angle, M, curEnergy, curAmmo);
                return 0;
            }
        case 'D':
            if ((M * 20) >= curAmmo) {
                fallBack(angle, M, curEnergy, curAmmo);
                return 0;
            }
            else {
                behaviour = 'G';
                break;
            }
        case 'G':
            if (M == 0) {
                printf("STOP");
            }
            if (abs(angle) < 5) {
                int P = (int)fmin(20, curAmmo);
                printf("FIRE %i", P);
                return 0;
            }
            else if (angle >= 5) {
                printf("LEFT %i", findX(curEnergy));
                return 0;
            }
            else if (angle <= -5) {

                printf("RIGHT %i", findX(curEnergy));
                return 0;
            }
        case 'P':
            if (M != 0) {
                behaviour = 'D';
                break;
            }
            else {
                if (offset >= -180 && offset <= -160) {
                    printf("BACKWARD %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > -160 && offset < -90) {
                    printf("RIGHT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= -90 && offset < -20) {
                    printf("LEFT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= -20 && offset <= 20) {
                    printf("FRONT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > 20 && offset <= 90) {
                    printf("RIGHT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset > 90 && offset < 160) {
                    printf("LEFT %i", findX(curEnergy));
                    return 0;
                }
                else if (offset >= 160 && offset <= 180) {
                    printf("BACKWARD %i", findX(curEnergy));
                    return 0;
                }
            }
        }
    }


    return 0;
}

void fallBack(int angle, int M, int curEnergy, int curAmmo)
{
    if (abs(angle) >= 5 || M == 0) {
        printf("BACKWARD %i", findX(curEnergy));
    }
    else {
        int P = (int)fmin(20, curAmmo);
        printf("FIRE %i", P);
    }
}
void advance(int angle, int M, int curEnergy, int curAmmo)
{
    if (abs(angle) >= 10 || M == 0) {
        printf("FRONT %i", findX(curEnergy));
    }
    else {
        int P = (int)fmin(20, curAmmo);
        printf("FIRE %i", P);
    }
}
int findX(int curEnergy)
{
    int X;
    if (curEnergy > 100) {
        X = 100;
    }
    else {
        X = curEnergy;
    }
    return X;
}

Event simple A+B get's around 900KB for me, but I've seen some ACs with 400KB in Java 1.8. Can you tell me how?

O(n+m) solution exists: extend <https://en.wikipedia.org/wiki/Tarjan%27s_off-line_lowest_common_ancestors_algorithm> to keep not only the top of each set in the disjoint-set forest, but also the weight of the path till that top.

Write code that way:

    if(n>=1000)
        puts("legion");
    else if (n>=500)
        puts("zounds");
    else if (n>=250)
        puts("swarm");
    else if (n>=100)
        puts("throng");
    else if (n>=50)
        puts("horde");
    else if (n>=20)
        puts("lots");
    else if (n>=10)
        puts("pack");
    else if (n>=5)
        puts("several");
    else puts("few");

use printf and scanf instead of cin,cout;

a=[]
for i in range(int(input())):
    c=str(input())
    if len(a)==0 :
        a.append(c)
        continue
    d=(a[0].split())[1]
    d=int(d)
    k=int((c.split())[1])

    if k>=d :
        a.reverse()
        a.append(c)
        a.reverse()
    else:
        for j in range (len(a)):
            h=int((a[j].split())[1])



            if k >= h:
                p = a[:j]
                o = a[j:]
                p.append(c)
                p.extend(o)








for i in range(len(a)):
   print(a[i])


Edited by author 26.09.2018 20:38

the number of integer point inside circle  is sigma(r*r/(4*i+1)-r*r/(4*i+3))
use stern borcot to cut hyperbola curve it can be done O(r^(2/3))

int main()
{
    optimize();
    int n,k;
    cin>>n>>k;
    k--;
    int z=k;
    bool b=0;
    int a[n+10];
    for(int i=0;i<n;i++)
    {
        if(k!=0)
        {
            if(b==0)
            {
                a[i]=k;
                b=1;
                z--;
            }
            else
            {
                if(z==0)
                {
                    a[i]=0;
                    k=0;
                    b=1;
                    continue;
                }
                a[i]=-k;
                k--;
                b=0;
                z--;
                if(z==0)
                    k=0;
            }
        }
        else
            a[i]=0;
    //dbg(z);
    }
    for(int i=n-1;i>=0;i--)
        cout<<a[i]<<" ";
    //return main();
}

when going to print 0 0 -3 3  or  -3 3 0 0 I am getting WA..But When I print 0 0 -1 1 I get AC..Why?what is the difference between 1st two case and last one??Can anyone explain me??

Thanks to Oleksandr Kulkov for preparing!

why is  k =10 and n =2 output =90

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main() {
unsigned long long k[4];
for(int i=0;i<4;i++){
    scanf("%llu",&k[i]);

}
printf("\n");
printf("%.4f\n",sqrt(k[3]));
printf("%.4f\n",sqrt(k[2]));
printf("%.4f\n",sqrt(k[1]));
printf("%.4f\n",sqrt(k[0]));
  return 0;
}

Ошибка на тесте 2 /Error in test 2

Edited by author 09.09.2018 15:45

Почините алгоритм проверки! Не контачит на сях. Вот код:
#include <stdio.h>

int main(int argc, char const *argv[])
{
    int a, b;
    a=1, b=5;
    printf("%i\n", a+b);
    return 0;
}

Hello, everyone.
Just wanted to summarize some info about this problem.
Ok, my code got AC with following things in it:
1. My pi was 3.1415926535897932384626433 (perhaps it is enough)
2. I used this formula from wikipedia
deltaArc=acos(sin(phiA)*sin(phiB)+cos(phiA)*cos(phiB)*cos(deltaL));
distance = deltaArc*3437.5;
(see the first formula from wikipedia http://en.wikipedia.org/wiki/Great-circle_distance)
3. I used the following condition:
if(100.00-distance>0.005) printf("DANGER!\n");
Hope it will help somebody.

1)Solution always exists and his complexity <= O(n^2)
2)Try to solve the problem when the permutation is n (n-1) ... (n-left+1) 1 2 3 ... right where
left+right=n.
   After O(n^2) steps you can get n 1 2 3 ... (n-1).
   Then after O(n^2) steps you can get 1 2 3 ... n.
3)Try to reduce the problem to this permutation by O(n^2) steps.
4)Good to know: 1 2 3 ... t x-> 2 1 3 ... t x -> 3 1 2 4 ... t x -> 4 1 2 3 5 ... t x ->... ->
t 1 2 3 ... (t-1) x -> x 1 2 3 ... t by t steps. I call this operation "Shift(t)".

Edited by author 26.07.2018 20:20

Edited by author 22.09.2018 14:02