Hi,I am Akanesasu.

I tried to solve 804F - Fake bullions several days ago,and i have a question about the analysis and the standing solutions.

The Link is above,and you can read the problem by yourself.

I have no doubt about the first part of the analysis,about how to calculate the Mn and Mx.

My question is that since we has already know the Mn and Mx of every gang,assume the A-th biggest Mn is E,we can just count the number of gangs whose Mx is equal to or greater than E.

If we assume this number is D,the answer should be C(D, B),where C is the number of combinations.

Because we can choose any B gangs from these D gangs,and there is always a way to set the score to satisfy the situation.

You may think thats not possible,but if we set the score of the gangs which we choose to the lowest score it can reach and at least E, and set the other score to the minimum,we can do it.

Prove:

The gangs whose Mn is greater than E has already considered before,the number of them is no more than A - 1. And the other gang's socre is just equal to or less than E,so the gangs we choose are all among the top gangs.

Look at this case:

5 2 2
01011
00011
11000
00100
00110
10 1110100101
1 1
7 1001000
4 1001
10 0110001100

As the whole graph is an SCC ,the mn and mx are:

mn[1]=6,mx[1]=10
mn[2]=1,mx[2]=1
mn[3]=2,mx[3]=7
mn[4]=2,mx[4]=4
mn[5]=4,mx[5]=10

We find that D is 4.

And if we set the scores as follow:

score[1]=6
score[2]=1
score[3]=4
score[4]=4
score[5]=4

then we can choose any 2 gangs from gang 1,3,4,5, so the answer should be 6,which is C(4,2).

But the standing code gives 4.

There is my submission:27191728

I wonder if the data and the analysis is incorrectly.Thanks!