# Friends Of Appropriate Ages（数学逻辑

Some people will make friend requests. The list of their ages is given and `ages[i]` is the age of the ith person.

Person A will NOT friend request person B (B != A) if any of the following conditions are true:

* `age[B]`

  `<`

  `= 0.5 * age[A] + 7`
* `age[B]`

  `>`

  `age[A]`
* `age[B]`

  `>`

  `100`

  `&`

  `&`

  `age[A]`

  `<`

  `100`

Otherwise, A will friend request B.

Note that if A requests B, B does not necessarily request A. Also, people will not friend request themselves.

How many total friend requests are made?

**Example 1:**

```
Input: [16,16]
Output: 2
Explanation: 2 people friend request each other.
```

**Example 2:**

```
Input: [16,17,18]
Output: 2
Explanation: Friend requests are made 17 -> 16, 18 -> 17.
```

**Example 3:**

```
Input: [20,30,100,110,120]
Output: 
Explanation: Friend requests are made 110 -> 100, 120 -> 110, 120 -> 100.
```

Notes:

```
1 <= ages.length <= 20000.
1 <= ages[i] <= 120.
```

分析

方法1 年级120内，map存每个age的个数。然后2个for遍历所有数字，相乘（注意= 是\*cnt(b)-1)

方法2 用array和和数学公式。a,b 之间范围的和相减，\[a, b = a/2 +7]

```
class Solution {
     public int numFriendRequests(int[] ages) {

        int map[] = new int[121];
        for(int age: ages){
            map[age] ++;
        }
        int ret = 0;
        for(int i = 1; i <= 120; i ++){
            for(int j = 1; j <= 120; j ++){//次数需要从1开始而不是从i， 因为朋友关系是相互的。所以需要前去后来
                if(valid(i,j)){
                    ret += map[i] * (i == j ? map[j] - 1 : map[j]);
                }
            }

        }
        return ret;
    }
    public boolean valid(int a, int b) {
        return !(b <= 0.5 * a + 7 || b > a || (b > 100 && a < 100)); //其实条件2和3重复了
    }
}
```

```
class Solution {
    public int numFriendRequests(int[] ages) { //优化2，另一数组sums记录范围，这样计算count不用2 for，直接找范围内个数即可
        int[] nums = new int[121], sums = new int[121];
        for (int age : ages) nums[age]++; //记录每个年龄多少人
        for (int i = 1; i < sums.length; i++) sums[i] = sums[i - 1] + nums[i]; //相当于记录小于i的有多少人
        int res = 0;
        for (int i = 15; i < sums.length; i++) { //i / 2 + 7 < i -> i>14
            if (nums[i] == 0) continue; //0一定要跳过，否则后面是负数
            int count = sums[i] - sums[i / 2 + 7]; //(i/2+7, i] 有多少个
            res += (count - 1) * nums[i]; //不能和自身request
        }
        return res;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://nataliekung.gitbook.io/ladder_code/facebook/friends-of-appropriate-agesff08-shu-xue-luo-ji.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.