题目
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?

这个题目特别有趣。最傻的办法就是来一个n^2循环每个pair相互用给出的三条规则比一下，符合规则requests的数量就+1。但是用屁股想也知道会TLE。

另外一个办法是，用一个size为121的array criterias
criterias[i] 表示有多少个人会对age为i的人make friend requests
你先把整个array扫一遍，就能填好criterias

接下来，再把array扫一遍，requests += criterias[ages[i]]
有一些corner case要注意: for age <= 14, none of the three criterias will satisfy
答案

class Solution {
    public int numFriendRequests(int[] ages) {
        int[] criterias = new int[121];
        int ans = 0, cnt = 0;
        for(int age : ages) {
            if(age <= 14) continue;
            double start = 0.5 * age + 7;
            if(start % 1 == 0) start += 1.0;
            else start = Math.ceil(start);
            for(int i = (int)start; i <= age; i++)
                criterias[i]++;
            cnt++;
        }

        for(int age : ages)
            ans += criterias[age];
        return ans - cnt;
    }
}