codeforces 320c A Problem about Polyline(数学)
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题目连接:http://codeforces.com/contest/579/problem/C
题意:给你一段波浪线,点数变化情况是—— (0,0)->(x,x)->(2x,0)->(3x,x),再给你一个点(a,b),让你输出尽可能小的x,使得(a,b)在这样的折线上。
思路:两段折线的方程分别是a-2kx=b,2kx-a=b。对于第一个式子,k=(a-b)/(2x)<=(a-b)/(2b),x=(a-b)/(2*k),其中k为整数,x为实数。用同样的方法对第二个式子求解,然后两者的x取较小的。
代码:
#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include<malloc.h>#include <map>#include <stack>#include <queue>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#include <limits.h>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-4)#define inf 0x3f3f3f3f#define sqr(x) (x) * (x)using namespace std;int main() { int a = 0, b = 0; scanf("%d %d", &a, &b); if (a < b) { printf("-1"); return 0;} printf("%.9lf", ( a + b ) / ( 2.0 * (( a + b ) / ( 2 * b )))); return 0;}
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