down

Try this test:
1 1 2 0
1 -1 0 1

Answer: 1 1 2 0

error in path recovery at a fixed level

for
2 = 2
3 = 12
4 = 104

can u give some tests?

Edited by author 21.10.2012 12:05

011

ans:
11

maybe forgot some reverse

3 1
2 1
2
0 1

you forgot about something

It is something wrong. Can somebody help me?

5 7 3 4
1 2
2 1
2 4
3 1
3 5
5 5
5 4
Answer: 0

I have two different solutions which handle this test very differently:

20 5
24
0.0 - 1.0
0.1 - 1.0
0.2 - 1.1
0.3 - 1.2
0.4 - 1.3
0.5 - 1.4
0.6 - 1.5
0.7 - 1.6
0.8 - 1.7
0.9 - 1.8
0.10 - 1.9
0.11 - 1.10
0.12 - 1.11
0.13 - 1.12
0.14 - 1.13
0.15 - 1.14
0.16 - 1.15
0.17 - 1.16
0.18 - 1.17
0.18 - 2.0
0.18 - 3.0
0.18 - 4.0
0.18 - 5.0
0.19 - 1.18
0 19

One of them might be even close to TLE, but both my solutions accepted here with time 0.015 s, which makes me think there is no such test in the list of tests.
I think this test should be added...

I don't understand, why did the authors create such a question that gets AC only if solved with Python or Java.

"...Она сразу же позвонила своей начальнице, затем своей подруге Лене, находящейся у неё в подчинении, потом Кате и Маше...". In my solution she can call friends, and then her boss. And solution, when she at first call the boss and then friends, get WA4.

...the(n) Masha...

I don't understand what dp has to do with this problem. I solved it using string hashing.

#include <stdio.h>
int main(void) {
    int a, b, c, max, min, med;
    scanf("%d %d %d", &a, &b, &c);
    max = a >= b && a >= c? a: b >= a && b >= c? b: c;
    min = a <= b && a <= c? a: b <= a && b <= c? b: c;
    med = a + b + c - min - max;
    if (med == 0)
        printf("%d", - max);
    else
        printf("%d", min - max * med);
    return 0;
}

I solved the problem the way I understood it but then when staring at the discussion page I discovered that buying the same tap twice is allowed.

Binary search the optimal value when only latin and spaces. You don't have to iterate through all m. Calculate transition for n in O(1) time. Overall complexity should be something like O(|s|log(m)).

If your code needs to calculate 10^i, then don't use BINPOW. Just precalculate. It wasted my time a little bit.