Apples


http://www.cnblogs.com/theskulls/p/5385898.html
果园里有一堆苹果,一共n头(n大于1小于9)熊来分,第一头为小东,它把苹果均分n份后,多出了一个,它扔掉了这一个,拿走了自己的一份苹果,接着第二头熊重复这一过程,即先均分n份,扔掉一个然后拿走一份,以此类推直到最后一头熊都是这样(最后一头熊扔掉后可以拿走0个,也算是n份均分)。问最初这堆苹果最少有多少个。
给定一个整数n,表示熊的个数,返回最初的苹果数。保证有解。
测试样例:
2
返回:3

    public int getInitial(int n) {
        // write code here
          for(int i=0;;i++){
 
            int apple = i*n+1; // 第一个人得到 i 个苹果,则第一个人分苹果时共有( i * n+1)个苹果  这个数要能 除以 n 余 1
            // 对第二个人分时候的苹果总数是: apple/n * (n - 1) 同样也要除以 n 余 1 ,下面循环判断  n个人都要满足这个条件。
 
            int t = apple;
 
            boolean flag = true;
 
            for(int j = n; j>0; j--){               
 
                if(t % n == 1 )
                    t = t/n * (n-1);// 第 i 个人分后的苹果数量  也就是 第 i + 1 人分之前的苹果数量 
 
                else{
                    flag = false;
                    break;
                }
 
            }
 
            if(flag) 
                return apple;
 
        }
    }
http://m.blog.csdn.net/article/details?id=51763820
设苹果总数为x,一共n头熊来分,均分成n份后多出一个,第i头熊扔掉多出的一个然后拿走自己的一份苹果,
直到最后一头熊(最后一头熊扔掉多出的一个后可以拿走0个)
1.第一头熊 苹果总数x,(x+n-1)能够被n整除
  分到的苹果 + 扔掉多出的一个苹果 = (x+n-1)/n 
2.第二头熊 苹果总数变为x-(x+n-1)/n
  分到的苹果 + 扔掉多出的一个苹果 = {[x - (x+n-1) / n] + (n - 1)}/n 
                               = Math.pow(n-1,1)(x+n-1)/Math.pow(n,2);
3.第三头熊 苹果总数变为x-(x+n-1)/n-(n-1)(x+n-1)/n*n
  分到的苹果 + 扔掉多出的一个苹果 = {[x - (x+n-1) / n - (n-1)(x+n-1)/n*n] + (n - 1)}/n 
                               = Math.pow(n-1,2)(x+n-1)/Math.pow(n,3);
...
4.第i头熊 
  分到的苹果 + 扔掉多出的一个苹果 = Math.pow(n-1,i-1)(x+n-1)/Math.pow(n,i);
5.最后一头熊
  分到的苹果 + 扔掉多出的一个苹果 = Math.pow(n-1,n-1)(x+n-1)/Math.pow(n,n);
  最后一头熊扔掉一个苹果,分到的苹果可以为0,则Math.pow(n-1,n-1)(x+n-1)/Math.pow(n,n)是正整数,即分子必须是分母的倍数
  因为(n-1)^(n-1) 与 n^n 互质,则有(x+n-1) = t*n^n,t取1时x最小,x = n^n-n+1;
    public int getInitial(int n) {
        int sum = (int)(Math.pow(n , n) - n + 1);
        return sum;
    }

题目解析: 这种带逻辑和数学的编程题在现在貌似很流行,这里采取的是数学归纳法的思想(根据多年做题的经验,这种题肯定是有表达式的),得出每一只熊得到的(分得的+扔掉的)苹果数,保证每一次苹果数是正整数。




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