第二题:
读题:
鸡兔同笼,n辆车,m个轮子,问电瓶车、三轮车、汽车各多少
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初级思路提供:
不要以为三层for暴力破解能过,我两层优化求汽车都过不了
参考代码:(TLE一个,最高复杂度25/6*1010)
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高级思路提供:
可以根据解不定方程的办法,把n和m作为常数考虑
先列出等量关系
然后根据等量关系1把k换做i和j表示
接着用等量关系2再把k换成i和j表示,以便于待会儿把j换做i表示
最后根据新得到的k=......的两个等量关系得到i和j两者等量关系
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代码:
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