【暑训排位 #6 B】 排列组合
N boxes are lined up in a sequence (1 ≤ N ≤ 20). You have A red balls and B blue balls (0 ≤ A ≤ 15, 0 ≤ B ≤ 15). The red balls (and the blue ones) are exactly the same. You can place the balls in the boxes. It is allowed to put in a box, balls of the two kinds, or only from one kind. You can also leave some of the boxes empty. It’s not necessary to place all the balls in the boxes. Write a program, which finds the number of different ways to place the balls in the boxes in the described way.
Input
Input contains one line with three integers N, A and B separated by space.
Output
The result of your program must be an integer written on the only line of output.
Example
input output
2 1 1
9
| 题意：a个红球，b个蓝球，放进n个桶里，桶可控，球可剩，问有多少种情况 |
思路：
红蓝两色分开讨论，然后相乘即可。对于每种颜色球，若有m个，我们分别讨论放0->m个球。这里就是一个隔板法
x 1 x_1 x1 + x 2 x_2 x2 + …+ x n x_n xn = 0 => 就1种情况
x 1 x_1 x1 + x 2 x_2 x2 + …+ x n x_n xn = 1 => 1-1个间隙 + n个可空位隔板=> C n + 1 − 1 1 C{^1_{n+1-1}} Cn+1−11
x 1 x_1 x1 + x 2 x_2 x2 + …+ x n x_n xn = 2=> 2-1个间隙 + n个可空位隔板=> C n + 2 − 1 1 C{^1_{n+2-1}} Cn+2−11
x 1 x_1 x1 + x 2 x_2 x2 + …+ x n x_n xn = 3=> 3-1个间隙 + n个可空位隔板=> C n + 3 − 1 1 C{^1_{n+3-1}} Cn+3−11
所以就累加1->m的情况数即可。
两个球情况累乘即是题要求输出。
AC代码：
#include