关于Problem8的改进

最大连续乘积

最新推荐文章于 2025-05-08 08:32:25 发布

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本文介绍了一种高效算法，用于寻找1000位数字中连续五个数字的最大乘积。通过四种不同的方法逐步优化，显著提高了计算效率。

问题描述

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

得到连续的五个数的最大乘积

思考：
每当我们得到当前五个数的乘积时，当挪动到下一个数，我们可以将保存的乘积乘以新的数除以之前五个数的第一个数。
如下
下标:0 1 2 3 4 5 6
数值:1 2 3 4 5 6 7
前五个数的乘积 result = 1*2*3*4*5
乘积中的第一个数first = 1
则求新的五个数的乘积 newResutl = result*6/first;
当然,实际运算时，我们还可以判断当前result和first是否为0

通过这样的运算，效率提高了很多~

自己逐步优化，写了四个方法
如下：

	public static int find_five_consecutive1() {
		int result = 1;
		int max = 0;
		String str = read();
		int begin = 0;
		int end = str.length();
		int step = 5;
		for (int i = begin; i < end - step; i++) {
			result = 1;
			for (int j = 0; j < step; j++) {
				result = result * Integer.parseInt(str.charAt(i + j) + "");
				count_for1++;
			}
			if (result > max) {
				max = result;
			}
		}
		return max;
	}

	public static int find_five_consecutive2() {
		int result = 1;
		int max = 0;
		String str = read();
		int begin = 0;
		int end = str.length();
		int step = 5;
		for (int i = begin; i < end - step; i++) {
			result = 1;
			for (int j = 0; j < step; j++) {
				if (str.charAt(i + j) == '0') {
					i = i + step + step - j;
					break;
				} else {
					int number = Integer.parseInt(str.charAt(i + j) + "");
					result = result * number;
					count_for2++;
				}
			}
			if (result > max) {
				max = result;
			}
		}
		return max;
	}

	public static int find_five_consecutive3() {
		int result = 0;
		String str = read();
		int step = 5;
		for (int i = 0; i < str.length(); i++) {
			if (str.charAt(i) == '0') {
				i = i + 5;
				count_for3++;
			} else {
				int multi = 1;
				for (int j = 0; j < step; j++) {
					count_for3++;
					char c = str.charAt(i + j);
					if (c == '0') {
						i = i + j + step;
						break;
					} else {
						multi *= Integer.valueOf(c + "");
						if (multi > result) {
							result = multi;
						}
					}
				}
			}
		}

		return result;
	}

	public static int find_five_consecutive4() {
		int result = 1;
		int max = 1;
		String str = read();
		int step = 5;
		int first = Integer.valueOf(str.charAt(0) + "");
		for (int i = 0; i < step; i++) {
			count_for4++;
			char c = str.charAt(i);
			result *= Integer.valueOf(c + "");
		}
		max = result;
		for (int i = 5; i < str.length(); i++) {
			if (result == 0) { // 重新计算
				result = 1;
				first = Integer.valueOf(str.charAt(i) + "");
				for (int j = 0; j < step; j++) {
					count_for4++;
					result *= Integer.valueOf(str.charAt(i + j) + "");
				}
				if(result>max){
					max = result;
				}
				i = i+step-1;
			} else{
				count_for4++;
				if(str.charAt(i)=='0'){
					result = 0;
					i = i+4;
					continue;
				}else{
					result = (result*Integer.valueOf(str.charAt(i)+""))/first;
					first = Integer.valueOf(str.charAt(i-4)+"");
					if(result>max){
						max = result;
					}
				}
			}
		}

		return max;
	}