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Re:是一个配方题

Posted by ghx at 2011-03-22 16:10:07 on Problem 1019
In Reply To:是一个配方题 Posted by:custqi at 2011-02-26 08:31:37

>    n<=i(i+1)/2 ->> 2*n+1/4<=(i+1/2)^2  ->> sqrt(2*n+1/4)<=i+1/2 ->> 
>    a = sqrt(2*n+1/4)-1/2<=i;
>       if a==floor(sqrt(2*n+1/4)-1/2) ->> ans = a;
>       else ans = n - i*(i+1)/2 {i=floor(sqrt(2*n+1/4)-1/2)}

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Re:是一个配方题 (28) ghx 2011-03-22 16:10:50

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