TopCoder Problem Statement - LCMRange |

Single Round Match 162 Round 1 - Division II, Level One |

Find the Least Common Multiple of a range of numbers.

01: public class LCMRange { 02: 03: public int lcm(int first, int last) { 04: 05: /* 06: * The Least Common Multiple must be a multiple of last, so we'll 07: * start there and increment by last to get the next value. 08: */ 09: int lcm = last; 10: 11: boolean found = false; 12: 13: // Continue until we find the LCM 14: while (!found) { 15: 16: int i = first; 17: 18: /* 19: * Check each number between first and last (inclusive) to ensure 20: * that lcm is a multiple. 21: */ 22: while (i <= last) { 23: 24: // If it doesn't divide evenly, skip this number. 25: if ((lcm % i) != 0) { 26: break; 27: } 28: 29: i ++; 30: } 31: 32: /* 33: * If we've tested all the numbers, then we've successfully found 34: * a Least Common Multiple 35: */ 36: if (i > last) { 37: found = true; 38: 39: } else { 40: 41: /* 42: * Otherwise try the next number. We know it must be a 43: * multiple of first, so we can increment by that amount. 44: */ 45: lcm += last; 46: } 47: } 48: 49: return lcm; 50: } 51: }

The constraints of this problem make it small enough that a brute-force method works well. We could just start at 1 and check to see if it is a Least Common Multiple (LCM) of each number in the range. If not, just increment by 1 and try again. A simple optimization is to note that our answer must be a multiple of last. So we can start our search there, and increment by the value of last to get the next number to test.

The % operator is used to determine if one number divides another evenly. if ((a % b) == 0) then b evenly divides a. Or, a is a multiple of b.

Note the break; statement in the inner-most while loop. This lets us quit on a potential LCM as soon as we find a number that it is not a multiple of. When the break is encountered, i will be <= last, and the if statement below will use this to determine if a LCM has been found or not.

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