UVA - 10970 Big Chocolate

Big Chocolate

Time Limit: 3000MSMemory Limit: Unknown64bit IO Format: %lld & %llu

Submit Status

Description

Problem G

Big Chocolate

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.

 

Figure 1. Mohammad’s chocolate

 

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

 

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Amirkabir University of Technology - Local Contest - Round #2

分析：

刘汝佳大白书里的超级大水题。本弱看到本题先是推了几种简单情况：如2*2,2*3,3*3，然后发现它们都满足ans=m*n-1.对于复杂的情况，肯定可以分解成这些简单情况，且切的时候也是同样的推理方式。于是果断撸下代码。1A了。

事后仔细想了想，发现一种更喜闻乐见的解释方式：每切一刀都能且只能多产生一块巧克力，一开始是1块巧克力，最后是m*n块巧克力，肯定切了m*n-1刀。

ac代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;

int main()
{
    int m,n;
    while(scanf("%d%d",&m,&n)==2)
    {
        printf("%d\n",m*n-1);
    }
    return 0;
}