TopCoder Problem Statement - PaperFold |

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

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

Calculate the minimum number of folds needed to fit a sheet of paper into a box.

01: public class PaperFold { 02: 03: public int numFolds(int[] paper, int[] box) { 04: 05: int paperMax = Math.max(paper[0], paper[1]); 06: int paperMin = Math.min(paper[0], paper[1]); 07: int boxMax = Math.max(box[0], box[1]); 08: int boxMin = Math.min(box[0], box[1]); 09: 10: int f1 = countFolds(paperMax, boxMax) + countFolds(paperMin, boxMin); 11: int f2 = countFolds(paperMax, boxMin) + countFolds(paperMin, boxMax); 12: 13: int folds = Math.min(f1, f2); 14: 15: return (folds <= 8) ? folds : -1; 16: } 17: 18: private static int countFolds(int a, int b) { 19: 20: int folds = 0; 21: 22: while (a > b) { 23: folds++; 24: b *= 2; 25: } 26: 27: return folds; 28: } 29: }

We need to consider two possibilities. First, that we can fold the paper length-wise and width-wise until it fits in box, and second, that we must first rotate the paper, and then fold it down to fit into the box. We'll try both ways, and go with the one that takes fewer folds.

On line 10, f1 is assigned the number of folds required if the maximum dimension of the paper is aligned with the maximum dimension of the box. On the next line, f2 is assigned the number of folds if this were rotated 90°.

The countFolds() method simply counts how many times the box's dimension needs to be doubled before it can fit the paper. I choose to double the size of the box rather than halving the size of the paper to avoid having to convert to doubles in order to deal with fractions.

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