#include <stdio.h>
int main(){
	int n;
	scanf ("%d",&n);
	while (n--){
		double m,x,y,z;
		scanf ("%lf%lf%lf%lf",&m,&x,&y,&z);
		printf ("%.2lf\n",(x*m/(y-x))*z);
	}
	return 0;
}

#include<stdio.h>
int main(){
    int n;
    scanf("%d",&n);
    while(n--){
        double m,x,y,z,s,a,b,c;
        scanf("%lf%lf%lf%lf",&m,&x,&y,&z);
        s=m*x;a=s/(z-x);b=s/(y-x)-a;c=(a+b)*z;
        printf("%.2f\n",c);
    }
}

#include<stdio.h>

int main(){

int n,m,y;

double x,z;

scanf("%d",&n);

while(n--){

scanf("%d %lf %d %lf",&m,&x,&y,&z);

printf("%.2f\n",(x*m)/(y-x)*z);

}

return 0;

}

经典华罗庚在火车上的题

from math import sqrt,ceil,gcd,log;re=lambda:map(int,input().strip().split())

t, = re()
for _ in range(t):
    m, x, y, z = re()
    print(f"{m * x * z / (y - x):.2f}")

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int n;
	cin>>n;
	int m,x,y,z;
	double t,s;
	while(n--)
	{
		cin>>m>>x>>y>>z;
		t = x*m*1.00/(y-x);
		s = t*z;
		cout<<fixed<<setprecision(2)<<s<<endl;
	}
}

和上面学长讲的一样

#include <stdio.h>

int main()

{

int n;

scanf("%d",&n);

for(int i=0;i<n;i++){

double m,x,y,z;

scanf("%lf %lf %lf %lf",&m,&x,&y,&z);

double dichu=m*x;

double c=0.0;

c=(dichu)/(y-x);

double k=c*z;

printf("%.2lf\n",k);

}

return 0;

}

//解题思路：求出哥哥追上弟弟的时间，再用时间乘以狗的速度。 #include <stdio.h> #include <stdlib.h>

int main() { int n; scanf("%d",&n); while(n--) { int m,x,y,z,t; double s; scanf("%d %d %d %d",&m,&x,&y,&z); if(y>x&&y<z)//限定条件 { t=(mx)/(y-x);//哥哥追上弟弟的时间 s=zt; printf("%.2lf\n",s); } } return 0; }