SRM146_DIV2

这次的题独立搞出来了前两道，可喜可贺

250

相同的数字加一起，找最大的和，因为是骰子，所以就是长度为6的数组

public class YahtzeeScore{    public int maxPoints(int[] toss){        int[] result = new int[6];        for (int i : toss){            result[i - 1] += i;        }        int max = 0;        for (int i : result){            max = Math.max(i, max);        }        return max;    }}

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500

长方体中的小长方体（不包括正方体）个数，找出所有的长宽组合就行~

public class RectangularGrid{    public long countRectangles(int width, int height){        long result = 0;        for (int i = 1; i <= width; i++){            for (int j = 1; j <= height; j++){                if ( i == j) continue;                result += (width + 1 - i) * (height + 1 - j);            }        }        return result;    }}

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1000

过河问题，这个题不会，看了别人的两种答案，一种是思考简单但是编程困难的穷举，一种是思考麻烦编程简单的方法

第一种是简单的递归调用吧

import java.util.Arrays;public class BridgeCrossing{    private boolean[] leftSide;    private boolean[] rightSide;    private int minTime = Integer.MIN_VALUE;    private int time = 0;    public int minTime(int[] times){        int length = times.length;        if (length == 1){            return times[0];        }        leftSide = newBooleanArray(length, true);        rightSide = newBooleanArray(length, false);        goToTheRight(times);        return minTime;    }    private boolean[] newBooleanArray(int length, boolean b){        boolean[] array = new boolean[length];        Arrays.fill(array, b);        return array;    }    private void goToTheRight(int[] times){        for( int i = 0; i < times.length - 1; i++){            if(!leftSide[i]) continue;            for (int j = i + 1; j < times.length; j++){                if(!leftSide[j]) continue;                // next state                leftSide[i] = leftSide[j] = false;                rightSide[i] = rightSide[j] = true;                int t = Math.max(times[i], times[j]);                time += t;                goToTheLeft(times);                // reset state                time -= t;                leftSide[i] = leftSide[j] = true;                rightSide[i] = rightSide[j] = false;            }        }    }    private void goToTheLeft(int[] times){        if(allTrue(rightSide)){            minTime = Math.min(minTime, time);        }else{            for( int i = 0; i < times.length; i++){                if(!rightSide[i]) continue;                // next state                rightSide[i] = false;                leftSide[i] = true;                time += times[i];                goToTheRight(times);                // rest state                time -= times[i];                rightSide[i] = true;                leftSide[i] = false;            }        }    }    private boolean allTrue(boolean[] rightSide){        for(boolean b : rightSide){            if(!b) return false;        }        return true;    }}

第二种，首先如果人数小于等于3就分情况讨论下，大于3则把行动最快的两个作为运送手电的桥梁，每完成一个轮回就会把两个目前最慢的人搬到对面。不停重复直到剩下的人数小于等于3。

import java.util.Arrays;public class BridgeCrossing{    public int minTime(int[] times){        int total = 0, remaining = times.length;        Arrays.sort(times);        while (remaining > 3){            total += Math.min(times[0] * 2 + times[remaining - 2] + times[remaining - 1], times[0] + times[1] * 2 + times[remaining - 1]);            remaining -= 2;        }        switch (remaining){            case 1:                return times[0];            case 2:                return total += times[1];            case 3:                return total += times[0] + times[1] + times[2];            default:                break;        }        return total;    }}