• ACM分巧克力问题


    Description

    Mohammad has recently visited Switzerland . As he loves his friends very much, he decided to buy some chocolate for them, but as this fine chocolate is very expensive(You know Mohammad is a little BIT stingy!), he could only afford buying one chocolate, albeit a very big one (part of it can be seen in figure 1) for all of them as a souvenir. Now, he wants to give each of his friends exactly one part of this chocolate and as he believes all human beings are equal (!), he wants to split it into equal parts.

    The chocolate is an rectangle constructed from  unit-sized squares. You can assume that Mohammad has also  friends waiting to receive their piece of chocolate.

    To split the chocolate, Mohammad can cut it in vertical or horizontal direction (through the lines that separate the squares). Then, he should do the same with each part separately until he reaches  unit size pieces of chocolate. Unfortunately, because he is a little lazy, he wants to use the minimum number of cuts required to accomplish this task.

    Your goal is to tell him the minimum number of cuts needed to split all of the chocolate squares apart.

     The Input

    The input consists of several test cases. In each line of input, there are two integers , the number of rows in the chocolate and , the number of columns in the chocolate. The input should be processed until end of file is encountered.

    The Output

    For each line of input, your program should produce one line of output containing an integer indicating the minimum number of cuts needed to split the entire chocolate into unit size pieces.

    Sample Input

    2 2

    1 1

    1 5

     

    Sample Output

    3

    0

    4

    解题思路:假设巧克力是一个M*N的,首先切(M-1)刀,把它分成M个大块,然后每一块都需要切(N-1)才能被切成小块。所以(M-1)+M*(N-1)=M*N-1;

    程序代码:

    #include <stdio.h>
    int main()
    {
          int M,N;
          while(scanf("%d%d",&M,&N)!=EOF)
                printf("%d
    ",N*M-1);

    return 0;
    }
  • 相关阅读:
    Oracle 创建索引分区
    Shiro SessionManager会话管理器设计概念
    管理表分区
    Oracle 表分区策略
    Oracle 表分区
    参考Shiro的Session定期验证会话失效的线程池
    HDU 3555 Bomb (数位dp)
    POJ 3666 Making the Grade (动态规划)
    Gym
    POJ
  • 原文地址:https://www.cnblogs.com/xinxiangqing/p/4655617.html
Copyright © 2020-2023  润新知