A. Berland Poker 、B. New Theatre Square

A. Berland Pokertime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

The game of Berland poker is played with a deck of nn cards, mm of which are jokers. kk players play this game (nn is divisible by kk).At the beginning of the game, each player takes nknk cards from the deck (so each card is taken by exactly one player). The player who has the maximum number of jokers is the winner, and he gets the number of points equal to x−yx−y, where xx is the number of jokers in the winner’s hand, and yy is the maximum number of jokers among all other players. If there are two or more players with maximum number of jokers, all of them are winners and they get 00 points.Here are some examples:n=8n=8, m=3m=3, k=2k=2. If one player gets 33 jokers and 11 plain card, and another player gets 00 jokers and 44 plain cards, then the first player is the winner and gets 3−0=33−0=3 points;n=4n=4, m=2m=2, k=4k=4. Two players get plain cards, and the other two players get jokers, so both of them are winners and get 00points;n=9n=9, m=6m=6, k=3k=3. If the first player gets 33 jokers, the second player gets 11 joker and 22 plain cards, and the third player gets 22 jokers and 11 plain card, then the first player is the winner, and he gets 3−2=13−2=1 point;n=42n=42, m=0m=0, k=7k=7. Since there are no jokers, everyone gets 00 jokers, everyone is a winner, and everyone gets 00 points.Given nn, mm and kk, calculate the maximum number of points a player can get for winning the game.InputThe first line of the input contains one integer tt (1≤t≤5001≤t≤500) — the number of test cases.Then the test cases follow. Each test case contains three integers nn, mm and kk (2≤n≤502≤n≤50, 0≤m≤n0≤m≤n, 2≤k≤n2≤k≤n, kk is a divisors of nn).OutputFor each test case, print one integer — the maximum number of points a player can get for winning the game.
Example
input
Copy
4
8 3 2
4 2 4
9 6 3
42 0 7
output
3
0
1
0
NoteTest cases of the example are described in the statement.

#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
LL t,n,m,sum,k;
int a[55];
int main()
{
      cin>>t;
      while(t--){
      cin>>n>>m>>k;
      if(m==0) cout<<0<<endl;
      else {
      sum=n-m;
      int num=n/k;
      if(m>num){
       int p=m-num;
       if(p%(k-1)==0) cout<<num-p/(k-1)<<endl;
       else cout<<num-p/(k-1)-1<<endl;
       }
       else cout<<m<<endl;
     }
   }
       return 0;
}

B. New Theatre Squaretime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

You might have remembered Theatre square from the problem 1A. Now it’s finally getting repaved.The square still has a rectangular shape of n×mn×m meters. However, the picture is about to get more complicated now. Let ai,jai,j be the jj-th square in the ii-th row of the pavement.
You are given the picture of the squares:
if ai,j=ai,j= “", then the jj-th square in the ii-th row should be black;
if ai,j=ai,j= “.”, then the jj-th square in the ii-th row should be white.
The black squares are paved already. You have to pave the white squares. There are two options for pavement tiles:
1×1 tiles — each tile costs x burles and covers exactly 1 square;
1×2 tiles — each tile costs y burles and covers exactly 2 adjacent squares of the same row.Note that you are not allowed to rotate these tiles or cut them into 1×1 tiles.
You should cover all the white squares, no two tiles should overlap and no black squares should be covered by tiles.What is the smallest total price of the tiles needed to cover all the white squares?InputThe first line contains a single integer tt (1≤t≤5001≤t≤500) — the number of testcases. Then the description of tt testcases follow.The first line of each testcase contains four integers nn, mm, xx and yy (1≤n≤100,1≤n≤100; 1≤m≤1000,1≤m≤1000; 1≤x,y≤10001≤x,y≤1000) — the size of the Theatre square, the price of the 1×11×1 tile and the price of the 1×21×2 tile.Each of the next nn lines contains mm characters. The jj-th character in the ii-th line is ai,jai,j. If ai,j=ai,j= "”, then the jj-th square in the ii-th row should be black, and if ai,j=ai,j= “.”, then the jj-th square in the ii-th row should be white.It’s guaranteed that the sum of n×mn×m over all testcases doesn’t exceed 105105.OutputFor each testcase print a single integer — the smallest total price of the tiles needed to cover all the white squares in burles.
Example
input
4
1 1 10 1
.
1 2 10 1
…
2 1 10 1
.
.
3 3 3 7
…*
…
..
output
10
1
20
18
Note
In the first testcase you are required to use a single 1×11×1 tile, even though 1×21×2 tile is cheaper. So the total price is 1010 burles.In the second testcase you can either use two 1×11×1 tiles and spend 2020 burles or use a single 1×21×2 tile and spend 11 burle. The second option is cheaper, thus the answer is 11.The third testcase shows that you can’t rotate 1×21×2 tiles. You still have to use two 1×11×1 tiles for the total price of 2020.In the fourth testcase the cheapest way is to use 1×11×1 tiles everywhere. The total cost is 6⋅3=186⋅3=18.

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <string>
#include<queue>
#include <cmath>
#define LL long long
#define PI 3.141592653589
using namespace std;
const LL maxn=1e9;
LL t,n,m,k,x,y;
char a[105][1005];
int main()
{
 cin>>t;
 while(t--){
  cin>>n>>m>>x>>y;
  memset(a,'*',sizeof a);
  for(int i=1;i<=n;i++)
   for(int j=1;j<=m;j++)
   cin>>a[i][j];
  int sum=0,num=0;
  for(int i=1;i<=n;i++)
   for(int j=1;j<=m;j++)
   {
    if(a[i][j]=='.'){
     if(a[i][j+1]=='.') {num++;a[i][j]='*';a[i][j]='*';j++;} 
    else sum++;
    }
   }
   if(y<=x) cout<<sum*x+num*y<<endl;
   else cout<<(sum+2*num)*x<<endl; 
 }
 return 0;
 }