I am working through the problems on project Euler and am not too certain if my understanding of the question is correct.
Problem 8 is as follows:
Only improvisation in my solution is, avoiding unnecessary computations by looking ahead.
package com.euler;
public class Euler8 {
public static void main(String[] ar) throws Exception {
String s =
"73167176531330624919225119674426574742355349194934" +
"96983520312774506326239578318016984801869478851843" +
"85861560789112949495459501737958331952853208805511" +
"12540698747158523863050715693290963295227443043557" +
"66896648950445244523161731856403098711121722383113" +
"62229893423380308135336276614282806444486645238749" +
"30358907296290491560440772390713810515859307960866" +
"70172427121883998797908792274921901699720888093776" +
"65727333001053367881220235421809751254540594752243" +
"52584907711670556013604839586446706324415722155397" +
"53697817977846174064955149290862569321978468622482" +
"83972241375657056057490261407972968652414535100474" +
"82166370484403199890008895243450658541227588666881" +
"16427171479924442928230863465674813919123162824586" +
"17866458359124566529476545682848912883142607690042" +
"24219022671055626321111109370544217506941658960408" +
"07198403850962455444362981230987879927244284909188" +
"84580156166097919133875499200524063689912560717606" +
"05886116467109405077541002256983155200055935729725" +
"71636269561882670428252483600823257530420752963450" ;
Integer[] tokens = new Integer[s.length()];
for (int i = 0; i < s.length(); i++) {
tokens[i] = (int) s.charAt(i)-48;
}
int prod = 1;
int[] numberSet = new int[5];
int prodCounter = 1;
for (int i=0; i<tokens.length-4; i++) {
// Look ahead: if they are zeros in next 5 numbers, just jump.
if ( tokens[i] == 0) {
i = i+1;
continue;
} else if ( tokens[i+1] == 0) {
i = i+2;
continue;
} else if ( tokens[i+2] == 0) {
i = i+3;
continue;
} else if ( tokens[i+3] == 0) {
i = i+4;
continue;
} else if ( tokens[i+4] == 0) {
i = i+5;
continue;
}
int localProd = tokens[i] * tokens[i+1] * tokens[i+2] * tokens[i+3] * tokens[i+4];
System.out.println("" + (prodCounter++) + ")" + tokens[i] + "*" + tokens[i+1] + "*" + tokens[i+2] + "*" + tokens[i+3] + "*" + tokens[i+4] + " = " + localProd);
if (localProd > prod) {
prod = localProd;
numberSet[0] = tokens[i];
numberSet[1] = tokens[i+1];
numberSet[2] = tokens[i+2];
numberSet[3] = tokens[i+3];
numberSet[4] = tokens[i+4];
}
}
System.out.println("Largest Prod = " + prod + " By: (" + numberSet[0] + " , " + numberSet[1] + " , " + numberSet[2] + " , " + numberSet[3] + " , " + numberSet[4] + ")");
}
}
Five single digits. 1, 5, 8... whatever shows up in the big number, all in a row. So if a chunk read "...47946285..." Then you could use "47946", "79462", "94628", "46285", etc.
A digit is a single 0-9 in the string representing the number. So the number 12345 has 5 digits. 1234554321 has 10 digits.
The product is the multiplicative total, not the added total. So the product of 3, 5 and 7 is 105.
A (somewhat clunky) way of rephrasing the question would be:
Given a 1000-digit number, select 5 consecutive digits from it that, when taken as individual numbers and multiplied together, give the largest result.