1，窮舉演算法

主要解決雞兔同籠類似問題

public class 窮舉演算法 {
	public static void main(String[] args) {
		int head = 35;
		int foot = 94;
		int j = 0;
		int i = 0;
		int tuziNumber = 0;
		int gongjiNumber = 0;
		for (i = 0; i <= 35; i++) {
			j = head - i;
			if (i * 2 + j * 4 == 94) {
				tuziNumber = i;
				gongjiNumber = j;
			}
		}
		System.out.println("兔子數量" + tuziNumber + "\n公雞數量" + gongjiNumber);
	}
}
 

2，遞推演算法

主要解決兔子產仔問題：一對兩個月大的兔子以後每個月可以生一對兔子，如果一月份出生，則三月份可產仔，求一年後有多少對兔子

思想：1-1,2-1,3-2,4-3,5-5......Fn=Fn-2+Fn-1

public class 遞推演算法 {
	public static int Fibonacci(int n) {
		if (n == 1 || n == 2) {
			return 1;
		} else {
			return Fibonacci(n - 1) + Fibonacci(n - 2);
		}
	}
	public static void main(String[] args) {
		int n = 12;
		System.out.println("兔子在" + n + "天，共有" + Fibonacci(n) + "只");
	}
}
 

3,分治演算法

主要解決硬幣真假類問題

思想：將一分二，二分四。。。。。

import java.util.*;

public class 分治演算法 {
	static final int MAXNUM = 10;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int coint[] = new int[MAXNUM];
		for (int i = 0; i < n; i++) {
			coint[i] = sc.nextInt();
		}
		int weizhi=FalseCoin(coint,0,n-1);
		System.out.println("假幣為"+weizhi+"號硬幣");
	}

	static int FalseCoin(int coin[], int low, int high) {
		int i, sum1, sum2, sum3;
		int re = 0;
		sum1 = sum2 = sum3 = 0;
		if (low + 1 == high) { // 兩枚硬幣
			if (coin[low] < coin[high]) {
				re = low + 1;			//防止索引為0，改為1號硬幣
				return re;
			} else {
				re = high + 1;
				return re;
			}
		}
		if ((high - low + 1) % 2 == 0) { // 如果是偶數個硬幣
			for (i = low; i <= low + (high - low) / 2; i++) { // 前半段硬幣
				sum1 = sum1 + coin[i]; // 求和
			}
			for (i = low + (high - low) / 2 + 1; i <= high; i++) { // 後半段硬幣
				sum2 = sum2 + coin[i]; // 求和
			}
			if (sum1 > sum2) {
				re = FalseCoin(coin, low + (high - low) / 2 + 1, high); // 後半段質量小，取後半段繼續二分求
				return re;
			} else {
				re = FalseCoin(coin, low, low + (high - low) / 2); // 前半段質量小，取前半段繼續二分求
				return re;
			}
		} else { // 如果是奇數個硬幣
			for (i = low; i <= low + (high - low) / 2 - 1; i++) {
				sum1 = sum1 + coin[i];
			}
			for (i = low + (high - low) / 2 + 1; i <= high; i++) {
				sum2 = sum2 + coin[i];
			}
			if (sum1 > sum2) {
				re = FalseCoin(coin, low + (high - low) / 2 + 1, high); // 後半段質量小，取後半段繼續二分求
				return re;
			} else if (sum1 < sum2) {
				re = FalseCoin(coin, low, low + (high - low) / 2); // 前半段質量小，取前半段繼續二分求
				return re;
			}
			if (sum1 == sum2) { // 如果是最中間的那個是假幣
				re = low + (high - low) / 2 + 1;
				return re;
			}
		}
		return re;
	}
}
 

4，概論演算法

主要求解Π的問題

思想：向正方形中撒撒豆子，求落在圓內的豆子，在比上總豆子，得到Π

public class 概論演算法 {
	public static void main(String[] args) {
		int count = 10000000;
		int number = 0;
		double PI;
		for (int i = 1; i < count; i++) {
			double x = Math.random();
			double y = Math.random();
			if (x * x + y * y <= 1) {
				number++;
			}
		}
		PI = 4.0 *number / count;
		System.out.println("Π的值為" +  PI);
	}
}