TC SRM 671 Div2 500 BearDartsDiv2

Problem Statement

Limak is an old brown bear who likes to play darts.

Limak just picked up an empty scorecard. He then threw a sequence of darts into a dartboard and for each dart he recorded the point value of the area it hit. You are given a int[] w containing the contents of Limak’s scorecard: a sequence of point values.

Today there is a special jackpot. In order to win the jackpot, the player must present a scorecard with exactly four scores: (a, b, c, d). Additionally, these scores must be such that a*b*c = d. Note that order matters: the scores a, b, c, d must have been obtained in this particular order.

Limak wants to erase all but four scores from his scorecard in such a way that he will win the jackpot. Compute and return the number of different ways in which he can do that.

Definition

Class BearDartsDiv2

Method count

Parameters vector

Returns long long

Method signature long long count(vector w)

(be sure your method is public)

Limits

Time limit (s) 1.000

Memory limit (MB) 256

Notes

Pay attention to the unusual time limit.

Constraints

w will contain between 4 and 500 elements, inclusive.

Each element in w will be between 1 and 10^6, inclusive.

Test cases

w { 10, 2, 2, 7, 40, 160 }

Returns 2

Limak must erase two of the six scores on his scorecard. There are two ways to do that and win the jackpot:

Erase the elements on 0-based indices 3 and 5. You are left with (a,b,c,d) = (10,2,2,40), and 10*2*2 = 40.

Erase the elements on 0-based indices 0 and 3. You are left with (a,b,c,d) = (2,2,40,160), and 2*2*40 = 160.

w { 128, 64, 32, 16, 8, 4, 2, 1 }

Returns 0

Regardless of which four scores Limak erases, the remaining four will always have a > b > c > d and therefore it cannot be the case that a*b*c = d.

w { 2, 3, 4, 5, 6, 8, 12, 16, 20, 24, 40, 24, 20, 16, 12, 8, 6, 5, 4, 3, 2 }

Returns 3

w { 100, 100, 100, 1000000, 1000000, 1000000, 1000000, 1000000 }

Returns 5

w { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }

Returns 2573031125

解析

meet in mid

把等式改写成a*b=d/c就变成O(N^2)的枚举啦！感谢@好地方bug的指导。可以写出如下代码（A就是题上的w）

map<LL,LL> cnt;
LL ans=0;
cnt[A[0]*A[1]]++;
for(int i=2;i<N-1;i++)//i枚举题中的c
{
for(int j=i+1;j<N;j++)
if(A[j]%A[i]==0)
ans+=cnt[A[j]/A[i]];
for(int j=0;j<i;j++)
cnt[A[j]*A[i]]++;
}

比赛的时候O(N^3)暴力乱搞，连最后一个样例都跑不过TAT。长者教导我还是要提高姿势水平……

而且我发现TC的时间好像测不准……上面的代码第一次T了，后来交又A了。