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题解：P5389 [Cnoi2019] 数学课

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比较目光呆滞的推柿子：
P(x=y)=∑v1,v2min⁡(v1v2)v1v2=∑i=1n∑j=1n3i(i+1)n(n+1)(n+2)3j(j+1)n(n+1)(n+2)min⁡(ai,aj)aiaj=∑i=1n∑j=1i3i(i+1)n(n+1)(n+2)3j(j+1)n(n+1)(n+2)1ai+∑i=1n∑j=i+1n3i(i+1)n(n+1)(n+2)3j(j+1)n(n+1)(n+2)1aj=∑i=1n∑j=1i3i(i+1)n(n+1)(n+2)3j(j+1)n(n+1)(n+2)2i(i+1)+∑i=1n∑j=i+1n3i(i+1)n(n+1)(n+2)3j(j+1)n(n+1)(n+2)2j(j+1)=∑i=1n∑j=1i6n(n+1)(n+2)3j(j+1)n(n+1)(n+2)+∑i=1n∑j=i+1n3i(i+1)n(n+1)(n+2)6n(n+1)(n+2)=∑i=1n∑j=1i6n(n+1)(n+2)3j(j+1)n(n+1)(n+2)+∑i=1n3i(i+1)(n−i)n(n+1)(n+2)6n(n+1)(n+2)=18n2(n+1)2(n+2)2(∑i=1n∑j=1ij(j+1)+∑i=1ni(i+1)(n−i))=18n2(n+1)2(n+2)2(∑i=1n∑j=1i(j2+j)+∑i=1n(−i3+(n−1)i2+ni))=18n2(n+1)2(n+2)2(∑i=1n(i(i+1)(i+2)3)+∑i=1n(−i3+(n−1)i2+ni))=18n2(n+1)2(n+2)2(13∑i=1n(i3+3i2+2i)+∑i=1n(−i3+(n−1)i2+ni))=18n2(n+1)2(n+2)2n2(n+1)2+2n(n+1)(2n+1)+4n(n+1)−3n2(n+1)2+2n(n−1)(n+1)(2n+1)+6n2(n+1)12=18n2(n+1)2(n+2)2−2n2(n+1)2+2n2(n+1)(2n+1)+4n(n+1)+6n2(n+1)12=18n2(n+1)2(n+2)2n(n+1)−2n(n+1)+2n(2n+1)+6n+412=18n2(n+1)2(n+2)2n(n+1)(2n2+6n+4)12=18n2(n+1)2(n+2)2n(n+1)2(n+2)6=3n(n+2)\begin{aligned} P(x=y)&=\sum\limits_{v_1,v_2}\dfrac{\min(v_1v_2)}{v_1v_2}\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{\min(a_i,a_j)}{a_ia_j}\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{1}{a_i}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{1}{a_j}\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{2}{i(i+1)}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}\dfrac{2}{j(j+1)}\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{6}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}+\sum\limits_{i=1}^n\sum\limits_{j=i+1}^n\dfrac{3i(i+1)}{n(n+1)(n+2)}\dfrac{6}{n(n+1)(n+2)}\\ &=\sum\limits_{i=1}^n\sum\limits_{j=1}^i\dfrac{6}{n(n+1)(n+2)}\dfrac{3j(j+1)}{n(n+1)(n+2)}+\sum\limits_{i=1}^n\dfrac{3i(i+1)(n-i)}{n(n+1)(n+2)}\dfrac{6}{n(n+1)(n+2)}\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^ij(j+1)+\sum\limits_{i=1}^ni(i+1)(n-i)\right)\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\sum\limits_{j=1}^i\left(j^2+j\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\sum\limits_{i=1}^n\left(\dfrac{i(i+1)(i+2)}{3}\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\left(\dfrac 1 3\sum\limits_{i=1}^n\left(i^3+3i^2+2i\right)+\sum\limits_{i=1}^n\left(-i^3+(n-1)i^2+ni\right)\right)\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n^2(n+1)^2+2n(n+1)(2n+1)+4n(n+1)-3n^2(n+1)^2+2n(n-1)(n+1)(2n+1)+6n^2(n+1)}{12}\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{-2n^2(n+1)^2+2n^2(n+1)(2n+1)+4n(n+1)+6n^2(n+1)}{12}\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)-2n(n+1)+2n(2n+1)+6n+4}{12}\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)\left(2n^2+6n+4\right)}{12}\\ &=\dfrac{18}{n^2(n+1)^2(n+2)^2}\dfrac{n(n+1)^2(n+2)}{6}\\ &=\dfrac{3}{n(n+2)}\\ \end{aligned}

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