leetcode-Array

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Ideas ：

1. nums If the number in can be divisible numsDivide All the numbers in , that , The condition it satisfies is division nunsDivide The least common divisor of
2. So we need to ask numsDivide The least common divisor of , You need to use rotation and multiplication

    int gcb(int a,int b){
    
        return a==0?b:gcb(b%a,a);
        // for instance , You can simply launch 
        // such as ,4 and 18 The least common divisor of , take 18 except 4 The remainder of （ by 2）, Then take the remainder 2 Keep following 4 Perform the same operation ,gcb（2,4）, Then you will find this time 4%2 by 0, So go back to 2, I.e. return to b
    }

3. The real thing about this question is （ After understanding the meaning of the topic, it turns into ：

1. loop nums Each number in , Find the smallest number that can be divided by the smallest common divisor k
2. Then traverse the loop again , This time I found everything better than k Small number

class Solution {
    
    public int minOperations(int[] nums, int[] numsDivide) {
    

    // The important point of this problem is to find the minimum common divisor of the rolling division 
    //nums If the number in can be divisible numsDivide All the numbers in , that , The condition it satisfies is division nunsDivide The least common divisor of 
    // So first use the rolling division method to find numsDivide The least common divisor of 
    int p=0;
    for(int numsdivide:numsDivide){
    
        p=gcb(p,numsdivide);
    }
    int res=Integer.MAX_VALUE;
    int sum=0;
    // Find the smallest divisible p Number of numbers 
    for(int i:nums){
    
        if(p%i==0){
    
            res=Math.min(res,i);
        }
    }
    if(res==Integer.MAX_VALUE)return -1;
    for(int i:nums){
    
            if(i<res)sum++;
    }
return sum;
    }

    int gcb(int a,int b){
    
        return a==0?b:gcb(b%a,a);
        // for instance , You can simply launch 
        // such as ,4  and 18 The least common divisor of , take 18 except 4 The remainder of , Then take the remainder and continue to follow 4 Perform the same operation , Until the remainder is 0, That's the top a=0, Just go back to b  such as 4%2 and 2 4%2 by 0 了 , Just go back to 2, I.e. return to b
    }
}