#include<iostream.h>
#include<math.h>
int prime[168]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,919,929,937,941,947,953,967,971,977,983,991,997};
int main()
{
    int i,j,k,m,n;
    cout<<prime[95];
    cin>>n;
    while(n)
    {
        cin>>m;
        if(m==0){cout<<0<<endl;goto next;}
        if(m==1){cout<<6<<endl;goto next;}
        for(i=0;i<168;i++)//长宽都是1
        if(m==prime[i]){cout<<m*4+2<<endl;goto next;}
        for(i=0;i<11;i++)//长是1
        if(m%prime[i]==0)
        {
            k=m/prime[i];for(j=0;j<95;j++)
            if(k==prime[j]){cout<<2*prime[i]*k+2*(prime[i]+k)<<endl;goto next;}
        }
        for(i=int(ceil(pow(m,1.0/3)));i>1;i--)//长宽高都不是1,至少长宽高有一个小于n的立方根，设为高
        if(m%i==0)
        {
            m=m/i;
            for(j=int(ceil(pow(m,0.5)));j>=2;j--)if(m%j==0)//高是1，长宽都不是1,至少长宽有一个小于n的平方根，设为长
            {k=m/j;cout<<2*(i*j+j*k+k*i)<<endl;goto next;}
        }
next:   n--;
    }
    return 0;
}

• 改进了很多，怎么还wa啊？郁闷啊！ (385) radish 2005-04-04 21:50:47
  • pow(m,1.0/3)外面加上ceil应该就好了 (427) frkstyc 2005-04-08 10:38:21
    • Re:pow(m,1.0/3)外面加上ceil应该就好了 (12) 00448217 2005-05-29 22:06:38
      • 求大于等于某个real number的最小整数 (14) frkstyc 2005-05-29 22:07:25
    • Re:漏了一种情况:n为两个相差不大的质数的乘积,如11*13 (4) mayp 2007-03-06 22:00:29
  • Re:600输出应为440 (2) sdfond 2008-03-04 19:30:18
• Re:高手帮我看看，怎么想也没发现问题，连0我都算了，怎么还wa啊？这个肯定不会超时的啦，因为我。。。 (1588) cxj666 2015-08-16 17:21:00
  • Re:高手帮我看看，怎么想也没发现问题，连0我都算了，怎么还wa啊？这个肯定不会超时的啦，因为我。。。 (80) cxj666 2015-08-16 17:25:21