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