title
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.
solution
累计每个age的count, 遍历age, count[i]*count[j]
static int Static_X = [](){
std::ios::sync_with_stdio(false);
cin.tie(NULL);
return 0;
}();
class Solution {
public:
int numFriendRequests(vector<int>& ages) {
int count[121] = { 0 };
for (auto it : ages) {
count[it]++;
}
int res = 0;
for (int i = 1; i <= 120; i++) {
for (int j = 1; j <= 120; j++) {
if (0.5*i+7>=j) continue;
if (i < j) continue;
if (j>100&&i<100) continue;
res += count[i] * count[j];
if (i == j)
res -= count[i];
}
}
return res;
}
};