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Re:我的代码In Reply To:我的代码 Posted by:074100215 at 2009-12-11 18:28:09 > 我考虑的是b,c其中一个肯定不大于2*a,b和c相差越少越好,所以从2*a开始找能整除的,找到就是相差差最少的,也就是和最小的 > > #include <stdio.h> > #include <cmath> > > int main() > { > __int64 a,b,c; > while(scanf("%I64d",&a)!=EOF) > { > b=2*a; > while((a*b+1)%(b-a)!=0) b--; > c=(a*b+1)/(b-a); > c=b+c; > printf("%I64d\n",c); > } > return 0; > } Followed by: Post your reply here: |
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