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微分方程+二分+枚举

Posted by Chan_keyword at 2020-07-28 18:50:20 on Problem 3933
分别求出x方向,y方向速度-时间,位移-时间公式(用微分方程,结果见https://kns.cnki.net/KCMS/detail/detail.aspx?dbcode=CJFQ&dbname=CJFD2008&filename=DXWL200812009&v=MDMwMzExVDNxVHJXTTFGckNVUjdxZllPUm5GeUhnVnIvSUlUWGNZckc0SHRuTnJZOUZiWVI4ZVgxTHV4WVM3RGg=)

因为theta是整数,所以枚举theta在(α,90]区间,对每一个θ,在时间区间(0,inf)间二分,
二分依据是斜率k=y/x与tan(alpha),
    if(k<tan(alpha))low = mid;
    else high = mid;
记录下最大x对应的theta即可。

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