HDU5874

Friends and Enemies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 171    Accepted Submission(s): 100

Problem Description

On an isolated island, lived some dwarves. A king (not a dwarf) ruled the island and the seas nearby, there are abundant cobblestones of varying colors on the island. Every two dwarves on the island are either friends or enemies. One day, the king demanded that each dwarf on the island (not including the king himself, of course) wear a stone necklace according to the following rules:
  
  For any two dwarves, if they are friends, at least one of the stones from each of their necklaces are of the same color; and if they are enemies, any two stones from each of their necklaces should be of different colors. Note that a necklace can be empty.
  
  Now, given the population and the number of colors of stones on the island, you are going to judge if it's possible for each dwarf to prepare himself a necklace.

 

Input

Multiple test cases, process till end of the input. 
  
  For each test case, the one and only line contains 2 positive integers M,N (M,N<231) representing the total number of dwarves (not including the king) and the number of colors of stones on the island.

 

Output

For each test case, The one and only line of output should contain a character indicating if it is possible to finish the king's assignment. Output ``T" (without quotes) if possible, ``F" (without quotes) otherwise.

 

Sample Input

20 100

 

Sample Output

T

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

最后竟成了找规律。。

1 0

2 1

3 2

4 4

5 6

6 9

7 12

8 16

9 20

10 25

11 30

12 36

13 42

......

 //2016.9.10
 #include <iostream>
 #include <cstdio>

 using namespace std;

 long long fun(int m)
 {
     long long ans;
     ans = (m/)*(m/);
     if(m%)ans += (m/);
     return ans;
 }

 int main()
 {
     long long n, m, res;
     while(scanf("%lld%lld", &m, &n)!=EOF)
     {
         res = fun(m);
         if(n>=res)
           printf("T\n");
         else printf("F\n");
     }

     return ;
 }