# [Bronze IV] King Arthur's Round Table [문제 링크](https://www.acmicpc.net/problem/34449) ## 문제 설명 <p>King Arthur has appointed you to find a table for he and his Knights of the Round Table to sit around while discussing the important matters of his kingdom.</p> <p>Sometimes, the knights can get into heated discussions where they passionately wave their arms around while debating one another. When this happens, King Arthur wants to make sure that each knight has enough room so as not to accidentally hit one another.</p> <p>As you search for tables, you need to ensure that the table's circumference is large enough to comfortably accommodate all of the knights, without risk of them accidentally hitting one another. King Arthur will tell you how many knights need to fit at the table as well as how much space he wants each knight to have. You have measured the diameter of each table. You must determine whether or not the table is suitable for King Arthur and his Knights of the Round Table.</p> ## 입력 <p>The input consists of three lines. The first line contains a single real number, $d$, representing the diameter of the table in feet. The next line contains a single real number, $w$, representing the width around the edge of the table required by each knight to sit comfortably in feet. The last line contains a single integer, $n$, representing the number of knights that need to fit around the table.</p> <p>Both $d$ and $w$ will be real numbers in the range $(0, 10000]$ and will have at most 9 digits after the decimal point. $n$ will be an integer between $1$ and $50\,000$, inclusive.</p> ## 출력 <p>Output <code>YES</code> if the table has enough room for all of King Arthur's Knights of the Round Table. Output <code>NO</code> if the table does <em>not</em> have enough room for all of King Arthur's Knights of the Round Table.</p> <p>Assume that $\pi = 3.14159$.</p> ## 풀이 원의 둘레 / 사람 수가 w 이상인지 확인하면 됩니다. ``` c++ #include<bits/stdc++.h> using namespace std; int main() { ios::sync_with_stdio(0); cin.tie(0); double d, w, n; cin >> d >> w >> n; cout << (d*3.14159/n>=w ? "YES" : "NO"); } ```