#include<stdio.h>
#include<math.h>
#define PI acos(-1)
int main()
{
    int N,i;
    double r;
    scanf("%d%lf",&N,&r);
    double xx,yy,cx,cy,fx,fy,sum=0,tsum=0,y,x,z;
    y=2*r;
    x=PI*r/2.0;
    z=N*(y-x);
    scanf("%lf%lf",&xx,&yy);
    fx=xx;
    fy=yy;
    for(i=2;i<=N;i++)
    {
        scanf("%lf%lf",&cx,&cy);
        sum=y+sum+sqrt((xx-cx)*(xx-cx)+(yy-cy)*(yy-cy));
        xx=cx;
        yy=cy;
        if(i==N)
        {
            sum=y+sum+sqrt((xx-fx)*(xx-fx)+(yy-fy)*(yy-fy));
            break;
        }
    }
    tsum=sum-z;
    printf("%.2lf\n",tsum);
    return 0;
}

i had that problem and solved it.
2 nd test is n=2 and answer is 0;

wa 23?

hello all,

I am new to programming problems.. I have a doubt regarding the solution.. I got the answer accepted but i want to know how did this problem come under the category of sorting??
I tried thinking a lot but i could never even guess that we have to use sorting.. I only came to know after i saw the hints in this forum.

SO please can anyone explain how did this problem come under sorting??

char buf[41];

unsigned hash()
{
    unsigned h = 0;
    for (char* p = buf; *p; ++p)
    {
        h = h * 13 + *p;
    }
    return h & 511;
}

Edited by author 22.04.2019 21:30

If you are using Edmonds-Karp, chnage it to regular Ford-Fulkerson (do DFS instead of BFS), because the running time in this case is better that way, as the maximum flow can be 1000 at most, so the complexity will be O(V*|f|) <= 2000*1.000.000 and this umber fits in 0.5 seconds. However, in case of E-K, the worst case running time is O(V^2*E) <= 1.000.000 * 1.000.000, which is the case of 1000-1000 complete graph and this exceeds the given 0.5 seconds.

Test
;;
+*
;+*;
Sample module returns:
Expression 1 evaluates to: 111
Expression 2 evaluates to: 1
Expression 3 evaluates to: 11

My AC program returns different result for 3rd string: 111

It's very weird that more complex expression returns smaller result, it fully contradicts the problem statement "It may be assumed that logic of expression evaluation does not depend on context. It means, that each subexpression is always evaluated in the same way with no dependency on it's entrance into the whole expression."

Well, I guess my test can be incorrect though, but it doesn't follow from the problem description. Maybe, it should be updated a little to specify what kind of expressions is acceptable?

well the problem is pretty simple and straight forward but when I try to solve this in python I tried out all algos like adding them that is N iteration then carrying them another N iteration. then print... or store them in a array then add from last digit that is also 2N iterations.... but is there any iteration which is less than 2N iteration including reading the values please help me!!

This question doesn't deserve difficulty above even 100.

sentenсe "for which x0 ≤ x, y0 ≤ y (as in the game the southeast wind blows)" looks like x0 <= x <= y and x0 <= y0 <=y but in fact it should be x0 <= x and y0 <= y

Try C = 794569207093795778
Answer: 315152584

[code deleted]

Edited by author 02.04.2019 21:28

Edited by author 02.04.2019 21:49

Edited by moderator 19.11.2019 23:34

Wonder why this task has a 147 complexity. Just sort and write

Factorize (in 1..9 figures) the N and then reduce the resulting number of figures.

I didn't understand if a player is hit by another ,what will be the changed condition of him? max(0,cur-1) or max(0,cur-2)?

n, k =map(int, input().split(' '))
z=0
time=0
num = 1
progress = 2
if k==1 and n==1:
    print(n)
elif k ==1:
    print(n-1)
elif k>=n:
    time+=1
    while num<n:
        num+=progress
        progress*=2
        time+=1
    print(time)
else:
    time+=1
    while z==0:
        if progress<=k and num<n:
            num+=progress
            progress*=2
            time+=1
        else:
            z+=1
    progress=k
    if num<n:
        while num<n:
            if num<n:
                time+=1
                num+=progress
    print(time)

Edited by author 16.04.2019 01:34

In the problem "2113. Survive the flood" I did't understand the condition of when the rabbit can jump to a cell and also I mean what's a jump height?

Edited by author 15.04.2019 18:51

Здравствуйте.
Это хорошо задача (для медианы не существует хорошего агрегата) требующая для точного решения хранения минимум N\2 чисел.

И тем более эффективная по времени решения, чем больше чисел мы одновременно можем хранить.

Поэтому хотелось бы точно знать что значит ограничение на 1MБ по памяти.

1. sort these orders for the following for loop;
2. can I serve current order? is there a not used day left for it? wow, that's easy if I can use std::set::upper_bound;
3. output our selected orders with delivery time.

// wow accepted, so lucky I am.

// after checked the discussion, found that this problem can be easily solve with priority queue, looks that priority queue is more convenient.