题干

思路

所有合金都需要由同一台机器制造，因此我们可以枚举使用哪一台机器来制造合金。

对于每一台机器，我们可以使用二分查找的方法找出最大的整数 xxx，使得我们可以使用这台机器制造 xxx 份合金。找出所有 xxx 中的最大值即为答案。

代码

class Solution {
    public int maxNumberOfAlloys(int n, int k, int budget, List<List<Integer>> composition, List<Integer> stock, List<Integer> cost) {
            // 全部转成 int[] 数组，效率比 List<Integer> 更高
	        int[] stocks = toArray(stock);
	        int[] costs = toArray(cost);
	        	
	        int ans = 0;
	        for(List<Integer> c : composition)
        		ans = calculate(budget, toArray(c), stocks, costs, ans);

    		return ans;
    	}
        //集合转换为数组
	    private int[] toArray(List<Integer> list) {
		       
		    int[] arr = new int[list.size()];
		    for(int i = 0; i < arr.length;i++)
		        arr[i] = list.get(i);

		    return arr;
    	}
        //检查
    	private boolean check(int budget, int[] compositions, int[] stocks, int[] costs, int count) {
    	    
    	    for(int i = 0; i < compositions.length; ++i) {
    	        int delta =  compositions[i] * count - stocks[i];
    	        if(delta <= 0) continue;
    	        
    	        budget -= delta * costs[i];
    	        if(budget < 0)
    	            return false;
		    }
    	  	
		    return true;
   	}
        //二分
        private int calculate(int budget, int[] compositions, int[] stocks,int[] costs, int count) {
        if(!check(budget, compositions, stocks, costs, count + 1))
            return count;
         
        count = count + 1;
        while(check(budget, compositions, stocks, costs, count << 1))
	        count <<= 1;
	         
	    int left = count,  right = count << 1;
	    while(left <= right) {
	        int mid = left + right >> 1;
	        if(check(budget, compositions, stocks, costs, mid)) 
	            left = mid + 1;
	        else 
	            right = mid - 1;
        }
         		
        return right;
	}	
}