div2 620 A

A

设所需要的时间为 \(t\),两只兔子之间的距离为 \(c ur\)

\(cut = (y-bt) - (x+at)\)

\(t = (y-x)/(a+b)\)

#include <bits/stdc++.h>
using namespace std;
int main() {
    int x,y,t,a,b,ans;
    cin >> t;
    while(t --) {
        cin >> x >> y >> a >> b;
        if((y - x) % (a + b) == 0) cout << (y - x) / (a + b) << endl;
        else cout << "-1\n";
    }
    return 0;
}

标签：620,cout,int,cin,ans,div2
来源： https://www.cnblogs.com/lukelmouse/p/12345160.html