pku acm 1183
来源:互联网 发布:java如何成为架构师 编辑:程序博客网 时间:2024/06/04 18:42
推导过程:
1/a = (1/b + 1/c)/ (1 - 1/(b*c)) => bc-1 = a(b+c)
assume b=a+m and c=a+n (b and c is always bigger than a)
(a+m)(a+n)-1=a(a+m+a+n) => a*a+a*n+a*m+m*n-1=2*a*a+m*a+n*a => m*n=a*a+1 and then
for(m=a;m>=1;m--)
if((a*a+1)%m==0) break;
n=(a*a+1)/m
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