HDU 5874 Friends and Enemies

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
找了很久的规律，终于确定，只有当n个人分成两个集合x和yx中的人互相敌对，y中的人互相敌对，并且x和y中的人互相友好时需要的颜色为x*y，而x+y=n,所以颜色要大于等于n/2*(n-n/2)才行。#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,LL>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-4;
const int INF = 0x7FFFFFFF;
const int mod = 998244353;
const int N = 1e5 + 10;
int n, m;

int main()
{
while (scanf("%d%d", &n, &m) != EOF)
{
printf("%s\n", 1LL * n * n / 4 <= m ? "T" : "F");
}
return 0;
}