The summer training of ZJU ICPC in July is about to end. To celebrate this great and happy day, the coaches of ZJU ICPC Team Navi and Fancy decided to BG everyone! Beside the two coaches, there are N participants. Those participants are divided into four groups to make problems for other groups.
After a brief discussion, they decided to go to Lou Wai Lou and have a great lunch. The cost for each person to have a meal in Lou Wai Lou is W yuan. Though it is very expensive, there is a special discount that Lou Wai Lou provided: for every K persons, one of them can have a free meal.
Navi and Fancy will equally pay all the cost for this great BG. Please help them to calculate how much money they each need to pay. By the way, please take care that the minimum monetary unit of RMB is fen (0.01 yuan).
Input
There are multiple test cases (no more than 100).
For each test case, there is only one line which contains three numbers N (1 <= N <= 100), W (0 <= W <= 10000) and K (1 < = K <= 100).
Output
For each test case, output the money a coach need to pay, rounded into 0.01 yuan.
Sample Input
3 70 3 32 999 1
Sample Output
140.00 0.00
Author: JIANG, Kai
Contest: ZOJ Monthly, March 2013
虽然说这是一道非常水非常水的题目，但是还是WA了我9次，应该说很多人被坑了= =
主要是要看懂这一句话 please take care that the minimum monetary unit of RMB is fen (0.01 yuan).
意思就是说分以后的要进位，例如3.461就要进位到3.47
一开始是没有看到两个教练平摊WA了，后来看到是输入lf觉得输入应该不止只有小数点后面两位，没想到还是贡献这么多WA。
#include<cstdio> #include<iostream> #include<cstring> #include<cmath> #include<stack> #include<queue> #include<stack> #include<stdlib.h> #include<algorithm> #define LL __int64 using namespace std; const double EPS=1e-8; int main() { int n,k; double w,ans; while(scanf("%d %lf %d",&n,&w,&k)!=EOF) { int num=(n+2)-(n+2)/k; ans=num*w/2*100; if(ans-(int)ans>EPS) ans=(int)ans+1; printf("%.2f ",ans/100); } return 0; }