2016 ACM/ICPC Asia Regional Dalian Online Friends and Enemies

本文链接：https://blog.youkuaiyun.com/qq_32866009/article/details/52496946

本文介绍了一个基于友谊与敌对关系的矮人项链问题，并提供了一种判断是否能根据岛上矮人的数量和石头颜色种类来为每个矮人分配符合特定规则的项链的方法。

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题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5874

Friends and Enemies

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 88 Accepted Submission(s): 50

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

下面是AC代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>

using namespace std;

int main()
{
    long long int m,n;
    while(~scanf("%I64d%I64d",&m,&n))
    {
        if(n>=(m*m)/4)
        {
            printf("T\n");
        }
        else
        {
            printf("F\n");
        }
    }
    return 0;
}