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Python解决codeforces ---- 1

13-10-08        来源:[db:作者]  
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 第一题 1A
A. Theatre Square
time limit per test 2 seconds
memory limit per test 64 megabytes
input standard input
output standard output
Theatre Square in the capital city of Berland has a rectangular shape with the size n × m meters. On the occasion of the city's anniversary, a decision was taken to pave the Square with square granite flagstones. Each flagstone is of the size a × a.
What is the least number of flagstones needed to pave the Square? It's allowed to cover the surface larger than the Theatre Square, but the Square has to be covered. It's not allowed to break the flagstones. The sides of flagstones should be parallel to the sides of the Square.
Input
The input contains three positive integer numbers in the first line: n,  m and a (1 ≤  n, m, a ≤ 109).
Output
Write the needed number of flagstones.
Sample test(s)
input
6 6 4
output
4
 
 
 题意:给一个n*m的矩形,和a*a的正方形。要用a*a的正方形去覆盖n*m的矩形,要求a*a的正方形不能切割开,但是两个a*a的正方形可以重叠,问最小需要几个这样的正方形
 代码:
[python]  
my_list = raw_input().split()  
  
n = int(my_list[0])  
m = int(my_list[1])  
a = int(my_list[2])  
  
print (n/a+(n%a>0))*(m/a+(m%a>0))  
 
 
 第二题 2A
A. Winner
time limit per test 1 second
memory limit per test 64 megabytes
input standard input
output standard output
The winner of the card game popular in Berland "Berlogging" is determined according to the following rules. If at the end of the game there is only one player with the maximum number of points, he is the winner. The situation becomes more difficult if the number of such players is more than one. During each round a player gains or loses a particular number of points. In the course of the game the number of points is registered in the line "name score", where name is a player's name, and score is the number of points gained in this round, which is an integer number. If score is negative, this means that the player has lost in the round. So, if two or more players have the maximum number of points (say, it equals to m) at the end of the game, than wins the one of them who scored at least m points first. Initially each player has 0 points. It's guaranteed that at the end of the game at least one player has a positive number of points.
Input
The first line contains an integer number n (1  ≤  n  ≤  1000), n is the number of rounds played. Then follow n lines, containing the information about the rounds in "name score" format in chronological order, where name is a string of lower-case Latin letters with the length from 1 to 32, and score is an integer number between -1000 and 1000, inclusive.
Output
Print the name of the winner.
Sample test(s)
input
3
mike 3
andrew 5
mike 2
output
andrew
input
3
andrew 3
andrew 2
mike 5
output
andrew
 
 题意:有很多人在玩纸牌游戏,总共玩n轮。每一轮里面包括玩家的名字,和玩家这一轮的得分,现在问最后谁的得分最高(假设为m),如果有多个人得分最高,输出最先得到m分的玩家
 思路:我们先利用n轮求出最后每个人的得分这样就可以求出最大的分数m,然后枚举这么多个人去找如果得分为m的人就去从头模拟一遍找到得分为m分的轮数,最后找到赢家
 代码:
[python]  
# define some variable  
n = int(raw_input())  
maxScore = {}  
input = []  
  
# n times input  
while n > 0:  
    list = raw_input().split()  
    input.append(list)  
    name = list[0]  
    score = int(list[1])  
    if maxScore.has_key(name):  
       maxScore[name] += score  
    else:  
       maxScore[name] = score  
    n -= 1  
  
# find maxScore = ans  
ans = 0  
for key in maxScore:  
    ans = max(ans , maxScore[key])  
  
# def to find the time >= ans  
def getTime(str):  
    sum = 0  
    cnt = 0  
    for list in input:  
        name = list[0]  
        score = int(list[1])  
        if name == str:  
           sum += score  
           if sum >= ans:  
              return cnt  
        cnt += 1   
  
# one by one if score == ans  
time = 2147483647  
for key in maxScore:  
    if maxScore[key] == ans:  
       t = getTime(key)  
       if time > t:  
          time = t  
          ansName = key  
  
# output  
print ansName  
  
        
 
 
 第三题 3A
A. Shortest path of the king
time limit per test 1 second
memory limit per test 64 megabytes
input standard input
output standard output
The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.
 
In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).
Input
The first line contains the chessboard coordinates of square s, the second line — of square t.
Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1to 8.
Output
In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: L, R, U, D, LU, LD, RU or RD.
L, R, U, D stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.
Sample test(s)
input
a8
h1
output
7
RD
RD
RD
RD
RD
RD
RD
 
 题意:给定一个8*8的地图,再给定两个点s和t,s是起点,t为终点,问s到t最少需要几步,规定每一步可以往当前格子的8个方向移动,最后打印出路径 
 思路:由于每一个点可以往周围的8个方向移动,那么我们很快的求出最少需要的步数为ans = max(abs(s.x-t.x) , abs(s.y-t.y)),我们求出了最少需要的几步之后我们直接去模拟,只要满足走这一步能够减少距离那么就是符合的走法
 代码:
[python] 
# Class for Point  
class Point:  
     def __init__(self , x , y):  
         self.x = x   
         self.y = y  
  
# input start and end Point  
str = raw_input()  
start = Point(8-int(str[1]) , ord(str[0])-97)  
  
str = raw_input()  
end = Point(8-int(str[1]) , ord(str[0])-97)  
  
# solve this problem  
dir = [[-1,0],[-1,1],[0,1],[1,1],[1,0],[1,-1],[0,-1],[-1,-1]]  
move = ["U","RU","R","RD","D","LD","L","LU"]  
ans = max(abs(start.x-end.x) , abs(start.y-end.y))  
  
# output  
print ans  
m = ans  
  
x = start.x  
y = start.y  
  
# print "%d %d" % (start.x , start.y)  
# print "%d %d" % (end.x , end.y)  
  
while m > 0:  
    for i in range(8):  
        tmpx = x+dir[i][0]  
        tmpy = y+dir[i][1]  
        if (tmpx >= 0 and tmpx < 8 and tmpy >= 0 and tmpy < 8):  
           dis = max(abs(tmpx-end.x) , abs(tmpy-end.y))    
           if dis < m:  
              print move[i]  
              x = tmpx  
              y = tmpy  
              break  
    m -= 1      
 
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