巴斯卡(Pascal)三角形基本上就是在解 nCr ,因為三角形上的每一個數字各對應一個nCr,其中 n 為 row,而 r 為 column,如下:
0C0
1C0 1C1
2C0 2C1 2C2
3C0 3C1 3C2 3C3
4C0 4C1 4C2 4C3 4C4
對應的數據如下圖所示:
解法
巴斯卡三角形中的 nCr 可以使用以下這個公式來計算,以避免階乘運算時的數值溢位:
nCr = [(n-r+1)/r] * nCr-1
nC0 = 1
演算法
/* 計算nCr,但是並不快,只是方便 */ Procedure COMBI(n, r) [ FOR(i = 1; i <= r; i = i + 1) p = p * (n-i+1) / i;
RETURN p; ]
解決 nCr 的算法之後,剩下的就是如何將這些數字排版成三角形的問題了,這就要看您是如何顯示成果的了,下面的程式將分別示範文字模式(C實作)與視窗模式(Java實作)的解法。
#include <stdio.h> #define N 12
long combi(int n, int r){ long p = 1; int i; for(i = 1; i <= r; i++) p = p * (n-i+1) / i; return p; }
void paint() { int n; for(n = 0; n <= N; n++) { int r; for(r = 0; r <= n; r++) { int i; // 排版設定開始 if(r == 0) { for(i = 0; i <= (N-n); i++) { printf(" "); } } else { printf(" "); } // 排版設定結束
printf("%3d", combi(n, r)); } printf("\n"); } }
int main() { paint(); return 0; }
import java.awt.*; import javax.swing.*;
public class Pascal extends JFrame { public Pascal() { setBackground(Color.white); setTitle("巴斯卡三角形"); setSize(520, 350); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); setVisible(true) }
private long combi(int n, int r){ int i; long p = 1;
for(i = 1; i <= r; i++) p = p * (n-i+1) / i; return p; }
public void paint(Graphics g) { final int N = 12; int n, r, t;
for(n = 0; n <= N; n++) { for(r = 0; r <= n; r++) g.drawString(" " + combi(n, r), (N-n)*20 + r * 40, n * 20 + 50); } }
public static void main(String args[]) { new Pascal(); } }
def combi(n, r): p = 1 for i in range(1, r + 1): p = p * (n - i + 1) / i return p
def paint(number): for n in range(number + 1): for r in range(n + 1): if r == 0: for i in range(number - n + 1): print(" ", end="") else: print(" ", end="") print("%3d" % combi(n, r), end="") print("") paint(12)
def combi(n: Int, r: Int) = { def c(p: Int, i: Int): Int = if(i <= r) c(p * (n-i+1)/i, i+1) else p c(1, 1) }
def paint(number: Int) { for(n <- 0 to number) { for(r <- 0 to n) { if(r == 0) for(i <- 0 to (number -n)) print(" ") else print(" ") printf("%3d", combi(n, r)) } println() } }
paint(12)
def combi(n, r) p = 1 1.upto(r) { |i| p = p * (n - i + 1) / i } p end
def paint(number) 0.upto(number) { |n| 0.upto(n) { |r| if r == 0 (number - n).times { print " " } else print " " end printf "%3d", combi(n, r) } puts } end
paint(12)
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