关于展开和式
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关于展开和式
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Vsinger_洛天依楼主2024/9/11 11:06

ans=ci=1n(ai+ci1ci)cS{0,1,2,,n}iS(aici)i∉Sci1ans=\sum\limits_{c}\prod\limits_{i=1}^{n}(a_i+c_{i-1}-c_i)\to \sum\limits_{c}\sum\limits_{S\subseteq \{0,1,2,\cdots,n\}}\prod\limits_{i\in S} (a_i-c_{i})\prod\limits_{i\not\in S}c_{i-1}

2024/9/11 11:06
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