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#include <bits/stdc++.h>
using namespace std;
/*
一批食物可供3个牛4个羊吃14天，或供4个牛15个羊吃7天，
请问这批食物可供6个牛与7个羊吃多少天？
(3x+4y)*14=(4x+15*y)*7
42x+56y=28x+105y
14x=49y
2x=7y
假设牛每天吃7,羊每天吃2可以求的假设总量（3*7+4*2）*14=29*14
根据6*7+7*2得出六头牛与7只羊每天吃草56
29*14/56=29/4=7.25天。
*/
int main() {
return 0;
}
