Description
Inhabitants of the Wonderland have decided to hold a regional programming contest. The Judging Committee has volunteered and has promised to organize the most honest contest ever. It was decided to connect computers for the contestants using a "star" topology - i.e. connect them all to a single central hub. To organize a truly honest contest, the Head of the Judging Committee has decreed to place all contestants evenly around the hub on an equal distance from it.Input
The first line of the input file contains two integer numb ers N and K, separated by a space. N (1 = N = 10000) is the number of cables in the stock, and K (1 = K = 10000) is the number of requested pieces. The first line is followed by N lines with one number per line, that specify the length of each cable in the stock in meters. All cables are at least 1 meter and at most 100 kilometers in length. All lengths in the input file are written with a centimeter precision, with exactly two digits after a decimal point.Output
Write to the output file the maximal length (in meters) of the pieces that Cable Master may cut from the cables in the stock to get the requested number of pieces. The number must be written with a centimeter precision, with exactly two digits after a decimal point.Sample Input
4 11 8.02 7.43 4.57 5.39
Sample Output
2.00
題意:有N條繩子,分別長Li,現將N條繩子分成K條長度相等繩子,求這K條繩子的最大可分割長度。
分析:想了一下,感覺一般方法都不太好用,關鍵點在於找出可以滿足K分割且臨界於繩子總長的的點,因此考到使用二分搜索,以100次二分精准查找到可行值,,,,POJ上C++可以通過,G++超時,貌似討論區沒人通過G++。。。。
#include#include #include const int MAXN=10000; const long INF=100000; double L[MAXN]; int N,K; int Tdfs(double x) { int num=0; for(int i=0;i =K; } int main() { int i; double lb=0,ub=INF,mid; while(std::cin>>N>>K) { for(i=0;i >L[i]; for(i=0;i<100;i++) { mid=(lb+ub)/2; if(Tdfs(mid)) lb=mid; else ub=mid; } printf("%.2f\n",floor(ub*100)/100); } return 0; }