题目描述
We have a long seat of width X centimeters. There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.
We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.
At most how many people can sit on the seat?
Constraints
All input values are integers.
1≤X,Y,Z≤105
Y+2Z≤X
输入
Input is given from Standard Input in the following format:
X Y Z
输出
Print the answer.
样例输入
13 3 1
样例输出
3
提示
There is just enough room for three, as shown below:
一道水题, 题意就是说给一个长度为x 公分的板凳,然后上面坐人,每个人占据 y 公分 ,间隔为z 公分 .
可以解方程 ,设可以最多坐 C 人 ,那么 x = Cy+z(C+1) ;
C = x-z/y+z ;
代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define d(x) for(int i = 1 ;i<=(x) ; i++)
#define Init_(x) memset(x,0,sizeof(x))
#define pi acos(-1)
#define while_(x) while(scanf("%d",&x)!=EOF)
#define while_c(x) while(x--)
#define scanf_(x,y) scanf("%d%d",&x,&y)
#define mst(a,b) memset((a),(b),sizeof(a))
using namespace std ;
typedef long long LL ;
const LL Mod = 1e9+7 ;
int main()
{
int x, y ,z ;
cin >> x >>y >>z ; // x 为长度 ,y每个人的宽度 ,z为间隔
int m = x-z ;
int n = y+z ;
cout<<m/n<<endl;
return 0 ;
}