Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5611 Accepted Submission(s): 2137

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<span style="font-size:14px;">#include<stdio.h>
#include<math.h>
int main()
{
int a,b;
while(~scanf("%d%d",&a,&b))
{
int lefta,leftb,righta,rightb,len1,len2,sum;
len1=ceil(sqrt(a));len2=ceil(sqrt(b));
lefta=(a-(len1-1)*(len1-1)-1)/2+1;
leftb=(b-(len2-1)*(len2-1)-1)/2+1;
righta=(len1*len1-a)/2+1;
rightb=(len2*len2-b)/2+1;
sum=fabs(len1-len2)+fabs(lefta-leftb)+fabs(righta-rightb);
printf("%d\n",sum);
}
return 0;
}
</span>原文地址:http://blog.csdn.net/ice_alone/article/details/38388397