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狂赞楼主~~,我再补充一些证明步骤In Reply To:原来是有规律的阿^_^ Posted by:noskill at 2005-03-27 10:12:23 ---------------------------------------------------------------------------------------- 1.1<a1 if a1=1, then a1(=1), a[t] together could be replaced by a[t]+1. reason: a[t]+1>a[t]*1 ---------------------------------------------------------------------------------------- 2.to all i, 1<=a[i+1]-a[i]<=2; if some i make a[i+1]-a[i]>2, then a[i],a[i+1] together could be replaced by a[i]+1,a[i+1]-1 together. reason: a[i]*a[i+1] < (a[i]+1)*(a[i+1]-1) (a[i]+1)*(a[i+1]-1)=a[i]*a[i+1]+a[i+1]-a[i]-1 so a[i+1]-a[i]-1>0 (* a[i+1]-a[i]>2) ---------------------------------------------------------------------------------------- 3. at MOST one i, fits a[i+1]-a[i]=2 if i<j and a[i+1]-a[i]=2 and a[j+1]-a[j]=2 then a[i],a[j+1] could be replaced by a[i]+1, a[j+1]-1 reason: a[i]*a[j+1]< (a[i]+1)*(a[j+1]-1) so a[j+1]-a[i]-1>0 (* a[j]-a[i]>=1 a[j+1]-a[j]>=1 so a[j+1]-a[i]>=2 ) ---------------------------------------------------------------------------------------- 4. a1<=3 if a1>=4, then a1,a2 together could be replaced by 2, a1-1, a2-1 together reason: a1*a2< 2*(a1-1)(a2-1) (a1-1)(a2-1)=a1*a2-a1-a2+1 so a1*a2>2*(a1+a2-1) (* a1>=4 and a2>=5) ---------------------------------------------------------------------------------------- 5. if a1=3 and one i fits a[i+1]-a[i]=2 then i must be t-1 if i<t-1 then a[i+2] could be replaced by 2, a[i+2]-2 together reason: a[i+2]<2*(a[i+2])-4 so a[i+2]>4 (* a[1]=3 a[3]>=5 so a[i+2]>=5) Followed by:
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