Description
Sewing Buttons with Grandma |
After so many years of studying math in the Academy of Colombian mathematics (ACM) in the Tropic, Eloi have Finall Y decided to visit he grandmother for the winter of the North hemisphere. "Buttons and patches and the cold wind blowing, the days pass quickly when I am sewing" She says-eloi now remembers Ho W Grandma Quilts has love in every stitch. As a matter of fact, he has decided to learn the useful handicraft art of sewing buttons with grandma.
Eloi have realized that there was an interesting mathematical puzzle related to the task of sewing buttons to the front of a Shirt. Given a collection ofn1 buttons of color C1, n2 buttons of Color c2, ..., nk buttons of color ck, and a shirt with m fron T buttonholes, in how many ways the buttonholes can is assigned m buttons from the n1 + ... + nk buttons?
Input
The input consists of several test cases. The first line of all test case contains, integersm and K separated by a blank (1m 50 ,1 k ). Thenk lines follow each containing a integer ni with 1 ni. The last test was followed by a line containing the zeros.
Output
The output consists of one line per test case containing the number of ways them buttonholes can be assigned m buttons from the n1 + ... + nk buttons, assuming that the colors c 1,..., ck is pairwise distinct.
Sample Input
1 33113 2423) 2110 0
Sample Output
370
Test instructions: give you m button position, then have n different colors of repeated buttons, to find the number of combinations
Idea: First try to assume that f[i][j] means that I have a different color of the buttons to make up J number of possible numbers, then
F [I][J]=< Span style= "Position:absolute; Top:-2.707em; Left:0.029em ">∑ i = 1 n < Span style= "Position:absolute; Top:-2.707em; Left:0.058em ">∑ j = 1 m ∑ k=0 a[i] F [I?1][J?k]?C [J][k]which c [ j [ k Represents the number of combinations
In the case of I-1, the K is picked from J to put the first I color, so multiply the number of combinations
Import Java.math.biginteger;import Java.util.scanner;public class Main {public static void main (string[] args) {final int MAXN = 105;int a[] = new INT[MAXN]; BigInteger f[][] = new BIGINTEGER[MAXN][MAXN]; BigInteger c[][] = new Biginteger[maxn][maxn];int N, m;for (int i = 0; I <=; i++) {c[i][0] = Biginteger.one;c[i][i] = Biginteger.one;} for (int i = 2; I <=, i++) for (int j = 1; j < I; j + +) C[i][j] = C[i-1][j-1].add (C[i-1][j]); Scanner in = new Scanner (system.in), while (true) {m = In.nextint (); n = in.nextint (); if (M = = 0 && n = = 0) break;for (int i = 1; I <= n; i++) A[i] = In.nextint (); for (int i = 0; I <=, i++) for (int j = 0; J <=; j + +) if (j = = 0 ) F[i][j] = Biginteger.one;else F[i][j] = biginteger.zero;for (int i = 1; I <= n; i++) for (int j = 1; j <= M; j + +) for (int k = 0; k <= a[i]; k++) if (j-k >= 0) F[i][j] = F[i][j].add (f[i-1][j-k].multiply (c[j][k))); System.out.println (F[n][m]);}}