#include <stdio.h>
int main()
{
int t1,t2,n;
int c=0;
scanf("%d %d", &t1, &t2);
for (int i=0; i<=9; i++)
{
    scanf("%d", &n);
    c+=(n*20);
}

t2-=c;
if (t2<t1)
{
   printf("Dirty debug :(");
}
else
{
    printf("No chance.");
}


}

Any help?

If you have WA 13, try this test:
1
a
1
b....ba , where b repeated ~130000 times

If you've got rte 27: m can be more than 10000

With help of my friend that was just looking at maximums index sequense and found an idea, finally, I got AC.
Some hints:
1) Find an idea to calculate next maximum indices from previous ones: all the maximum's indices can be obtained from previous ones in such ways:
i = s*2 + 1; or
i = s*2 - 1; or
i = s*4 + 1; or
i = s*4 - 1;

for example f(21)=8 is maximum for 21<=n<=34
so 21 = 2*11-1, where 11 is one of previous maximums indices. And so on using all the formulas above.
but 2*11+1 = 23 is not a maximum index. so we have to cancel  this number (23).
2) Learn to calculate f(n) in O(logn)
3) Generate all maximums indices (about 1500). You can store them in heap to get an access to the smallest maximum index.
4) than read n and just search for it in maximums index array.
P.S. who can prove the first and the main idea? If you can, please post here you proof.

You only have four things to pick from, so you can calculate each step explicitly. This equation follows the notation in the article: T is the time to collect all of the items, and ti be the time to collect the i-th item after i−1 items have been collected.

E(T)=E(t1)+E(t2)+E(t3)+E(t4)=p1−1+p2−1+p3−1+p4−1.
The probability p1 of picking a new one if you have picked none yet is 1.

The probability p2 of picking a new one if you have picked one depends on which one you picked first. Let's call the items A through D and the probabilities of picking each item if they're all in the box pa through pd.

Then p2 would be
pa(pb+pc+pd)+pb(pc+pd+pa)+pc(pd+pa+pb)+pd(pa+pb+pc).
The above expression is the probability of picking a new one given that A had been picked already, plus the probability of picking a new one given that B had been picked already, and so on.

The expression for p3 will have six terms in the sum; one of these will be papb(pc+pd). The expression for p4 will have four terms in the sum (actually, it's the same term four times!)

What can I do?.. My program passes all testes here, but still WA12... Any test cases?

it takes a lot of my times.

#include<iostream>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    cout << b-1 << ' ' << a - 1;
}

4
2 2 3
2 1 3
3 1 2 4
1 3

6
5 2 3 4 5 6
5 1 3 4 5 6
5 1 2 4 5 6
5 1 2 3 5 6
5 1 2 3 4 6
5 1 2 3 4 5

8
7 2 3 4 5 6 7 8
7 1 3 4 5 6 7 8
7 1 2 4 5 6 7 8
7 1 2 3 5 6 7 8
7 1 2 3 4 6 7 8
7 1 2 3 4 5 7 8
7 1 2 3 4 5 6 8
7 1 2 3 4 5 6 7

6
2 2 3
2 1 3
2 1 2
2 5 6
2 4 6
2 4 5

diameter of 6875 miles (/ 2 = 3437.5)???

the mean radius of earth is 6371.0 km (3958.8 miles)

a = input()
minussum=0
plussum=0
b =[0:3]
c =[3:6]
c=int(c)
minus = c-1
plus = c+1
minus=str(minus)
plus=str(plus)
for i in minus:
    i=int(i)
    minussum+=i
for i in plus:
    i=int(i)
    plussum+=i
n=0
for i in b:
    i=int(i)
    n+=i
if n == minussum or n == plussum:
    print("Yes")
else:
    print("No")




Help please)
Why re?

Edited by author 13.09.2022 23:21

Python files encoded in UTF-8 with BOM always get "Runtime Error 1".
It would be nice if it was fixed.

I just do maxsimum 30 iteration on my super all doing bruteforce and this give me right ans and i got ac

I could solve it using trivial bruteforce. But I get TL, using more clever and probably expected solution. I used this: https://e-maxx.ru/algo/discrete_root (Read only first section, we don't need anything else). The slowest part here is Discrete Logarithm, which I calculate in \sqrt{n} * log_from_map.

My program stuck at WA-17 for several times and I don't know the reason why. Can anyone give me some challenging test cases?

The more - the better! 'cause I have no idea what I'm doing wrong :\

#include <iostream>
using namespace std;

int main() {
    int leavingCars, minutes;
    int carPerMinute[100];
    int totalCars = 0;

    cin >> leavingCars >> minutes;
    int carsLeavingPerMinute = leavingCars * minutes;

    for(int index = 1; index <= minutes; ++index) {
        cin >> carPerMinute[index];
        totalCars += carPerMinute[index];
    }

    cout << totalCars - carsLeavingPerMinute;

    return 0;
}

Are you going to add javascript (especially nodejs) support for this online judge | Планируется ли добавить на сайт поддержку javascript или nodejs?