poj 1942 Paths on a Grid
2016-10-11 21:18
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Paths on a Grid
Description
Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b)2=a2+2ab+b2). So you decide to waste
your time with drawing modern art instead.
Fortunately you have a piece of squared paper and you choose a rectangle of size n*m on the paper. Let's call this rectangle together with the lines it contains a grid. Starting at the lower left corner of the grid, you move your pencil to the upper right corner,
taking care that it stays on the lines and moves only to the right or up. The result is shown on the left:
Really a masterpiece, isn't it? Repeating the procedure one more time, you arrive with the picture shown on the right. Now you wonder: how many different works of art can you produce?
Input
The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m, denoting the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension. Input is terminated by n=m=0.
Output
For each test case output on a line the number of different art works that can be generated using the procedure described above. That is, how many paths are there on a grid where each step of the path consists of moving one unit to the right or one unit up?
You may safely assume that this number fits into a 32-bit unsigned integer.
Sample Input
Sample Output
/*
给一个n*m的矩阵网格,问有多少种方法从左下角走到右上角。
注意n,m都是32位无符号整形范围内,再从左下角走到右上角的过程中,每次只能向上或向右走一个单位长度。
解题思路:
向上走有n中方法,向右走有m种方法,先把就相当于n+m个位置选n个或选m个(选了n个那m个就固定了,选m个同理剩下n个也固定了)。
所以代码主要实现的是如何求组合数
.....
*/
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#define MAX_N 1000005
using namespace std;
long long comp(long long n, long long m)
{
long long j, i = max(n, m)+1;
long long sum = 1;
for ( j = 1;j <= min(n, m); j++ )
{
sum = sum*i/j;
i++;
}
return sum;
}
int main()
{
long long n, m;
while ( scanf ( "%lld %lld", &n, &m )&&( n != 0|| m != 0 ) )
{
long long sum = comp(n ,m);
printf ( "%lld\n", sum );
}
}
代码菜鸟,如有错误,请多包涵!!!
如有帮助记得支持我一下,谢谢!!!
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 24985 | Accepted: 6219 |
Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b)2=a2+2ab+b2). So you decide to waste
your time with drawing modern art instead.
Fortunately you have a piece of squared paper and you choose a rectangle of size n*m on the paper. Let's call this rectangle together with the lines it contains a grid. Starting at the lower left corner of the grid, you move your pencil to the upper right corner,
taking care that it stays on the lines and moves only to the right or up. The result is shown on the left:
Really a masterpiece, isn't it? Repeating the procedure one more time, you arrive with the picture shown on the right. Now you wonder: how many different works of art can you produce?
Input
The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m, denoting the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension. Input is terminated by n=m=0.
Output
For each test case output on a line the number of different art works that can be generated using the procedure described above. That is, how many paths are there on a grid where each step of the path consists of moving one unit to the right or one unit up?
You may safely assume that this number fits into a 32-bit unsigned integer.
Sample Input
5 4 1 1 0 0
Sample Output
126 2
/*
给一个n*m的矩阵网格,问有多少种方法从左下角走到右上角。
注意n,m都是32位无符号整形范围内,再从左下角走到右上角的过程中,每次只能向上或向右走一个单位长度。
解题思路:
向上走有n中方法,向右走有m种方法,先把就相当于n+m个位置选n个或选m个(选了n个那m个就固定了,选m个同理剩下n个也固定了)。
所以代码主要实现的是如何求组合数
.....
*/
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <iostream>
#define MAX_N 1000005
using namespace std;
long long comp(long long n, long long m)
{
long long j, i = max(n, m)+1;
long long sum = 1;
for ( j = 1;j <= min(n, m); j++ )
{
sum = sum*i/j;
i++;
}
return sum;
}
int main()
{
long long n, m;
while ( scanf ( "%lld %lld", &n, &m )&&( n != 0|| m != 0 ) )
{
long long sum = comp(n ,m);
printf ( "%lld\n", sum );
}
}
代码菜鸟,如有错误,请多包涵!!!
如有帮助记得支持我一下,谢谢!!!
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