UVA 10970 第一次比赛 D题（后面才补的）

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 1 ≤ M ≤ 300, the
number of rows in the chocolate and 1 ≤ N ≤ 300, the number of columns in the chocolate. The input
should be processed until end of file is encountered.
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

题意：给你一块M*N的巧克力，问把它切成最小单元需要最少切几刀，分开的就不能一起切了。

思路：既然长，宽是N,M，那么就要分成N*M块，自然就需要M*N-1刀了（当初比赛的时候没有做出来，自己也是够蠢的）

代码如下：

 #include <stdio.h>

 int main()

 {

     int N,M;

     while(scanf("%d%d",&N,&M)==)

     {

         int ans=M*N-;

         printf("%d\n",ans);

     }

     return ;

 }

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UVA 10970 第一次比赛 D题（后面才补的）

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UVA 10970 第一次比赛 D题（后面才补的）