UVA10970 Big Chocolate水题

Posted 2021-02-10 tigerisland45

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 M × N rectangle constructed from M × N unit-sized squares. You can assume that
Mohammad has also M × N 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 M × N 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.
Input
The input consists of several test cases. In each line of input, there are two integers 1M300, thenumber of rows in the chocolate and 1N300, the number of columns in the chocolate. The inputshould be processed until end of le is encountered.
Output
For each line of input, your program should produce one line of output containing an integer indicatingthe 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

问题链接：UVA10970 Big Chocolate
问题简述：
????将一个mn的巧克力切成11的巧克力，最少需要切几刀？
问题分析：
????没有思路就从小到大枚举一下。11的话只需要切0刀；12或21的话只需要切1刀；22的话需要切3刀。mn的话需要切mn-1刀。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10970 Big Chocolate */

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int m, n;
    while(~scanf("%d%d", &m, &n))
        printf("%d
", m * n - 1);

    return 0;
}