zoukankan      html  css  js  c++  java
  • 比赛大巧克力问题解题报告

    大巧克力问题

    题目大意:

    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.

    要求:

    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.

     输入样例:

    Sample Input

    2 2

    1 1

    1 5

     

    Sample Output

    3

    0

    4

    题目分析:

    题目内容很多,但题目很简单,前面说了很多没用的东西。根据给出的样例分析,可以判断出一次只能将巧克力分成两块,所以切的次数a=M*N-1。

    程序代码:

     1 #include<cstdio>
     2 #include<iostream>
     3 using namespace std;
     4 
     5 int M[300];
     6 int N[300];
     7 
     8 int main()
     9 {
    10     int M,N,a;
    11     while(scanf("%d%d",&M,&N)!=EOF)
    12     {
    13         a=M*N-1;
    14         cout<<a<<endl;
    15     }
    16     return 0;
    17 }

    心得:

    这道题目很简单,但由于比赛心情很烦躁,所以没有好好看这道题。比赛完后再看这道题才发现这道题原来如此简单。这次比赛我看到了和别人之间的差距,我应该多做一些题目,好好看一下C语言的基础知识。下次比赛要调整好自己的心态。

  • 相关阅读:
    计算机基础总结
    Apache安装错误 APR not found解决方法
    一、编译错误
    2.2 进程控制之进程共享
    2.1 进程控制之fork创建子进程
    ARM串口控制终端命令
    u-boot、kernel、root系统烧写和挂载命令命令
    8.1 编写USB鼠标驱动程序,并测试
    八、USB驱动分析
    Source Insight的使用
  • 原文地址:https://www.cnblogs.com/ttmj865/p/4655768.html
Copyright © 2011-2022 走看看