projecteuler Problem 8 Largest product in a series
The four adjacent digits in the 1000-digit number that have the greatest product are 9 × 9 × 8 × 9 = 5832.
Find the thirteen adjacent digits in the 1000-digit number that have the greatest product. What is the value of this product?
譯文:
觀察著1000個數字,通過運算可得,最大的連續的四位數的乘機為5832,求出最大的連續的13位數的乘機?
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第一次code:
import java.util.ArrayList; import java.util.List; public class Main { public static void main(String[] args) { System.out.println(num(run())); } /* * 將1000個數放到一個字符串 * 將字符串轉換成字符串數組 * 計算每連續的13位數的乘機,並放置在Arraylist數組中 * * 注意:乘機的類型為long,如果為int的話,數據結果會超出,導致計算結果出錯 */ public static List run() { String a="73167176531330624919225119674426574742355349194934" +"96983520312774506326239578318016984801869478851843" +"85861560789112949495459501737958331952853208805511" +"12540698747158523863050715693290963295227443043557" +"66896648950445244523161731856403098711121722383113" +"62229893423380308135336276614282806444486645238749" +"30358907296290491560440772390713810515859307960866" +"70172427121883998797908792274921901699720888093776" +"65727333001053367881220235421809751254540594752243" +"52584907711670556013604839586446706324415722155397" +"53697817977846174064955149290862569321978468622482" +"83972241375657056057490261407972968652414535100474" +"82166370484403199890008895243450658541227588666881" +"16427171479924442928230863465674813919123162824586" +"17866458359124566529476545682848912883142607690042" +"24219022671055626321111109370544217506941658960408" +"07198403850962455444362981230987879927244284909188" +"84580156166097919133875499200524063689912560717606" +"05886116467109405077541002256983155200055935729725" +"71636269561882670428252483600823257530420752963450"; char [] b = a.toCharArray(); List<Long>list = new ArrayList<Long>(); for(int i=0;i<b.length-13;i++) { list.add(Long.valueOf(String.valueOf(b[i]))*Long.valueOf(String.valueOf(b[i+1])) *Long.valueOf(String.valueOf(b[i+2]))*Long.valueOf(String.valueOf(b[i+3])) *Long.valueOf(String.valueOf(b[i+4]))*Long.valueOf(String.valueOf(b[i+5])) *Long.valueOf(String.valueOf(b[i+6]))*Long.valueOf(String.valueOf(b[i+7])) *Long.valueOf(String.valueOf(b[i+8]))*Long.valueOf(String.valueOf(b[i+9])) *Long.valueOf(String.valueOf(b[i+10]))*Long.valueOf(String.valueOf(b[i+11])) *Long.valueOf(String.valueOf(b[i+12]))); } return list; } /* * 遍歷ArrayList數組 * 比較輸出最大值 */ public static long num(List list) { long max=(Long)list.get(0); for(int i=0;i<list.size();i++) { if(max < (Long)list.get(i)) { max = (Long)list.get(i); } } return max; } }
Answer : 23514624000
時間效率: 5 毫秒。
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