Joking with Fermat's Last Theorem UVA - 12665

AC代码（暴力就可以然后范围从输入的x到y*10+3的1/3次方因为c是最大的然后i*i*i+j*j*j%10==3&&<=y*10+3）

Select Code

#include<cstdio>
#include<cstring>
#include <bits/stdc++.h>
using namespace std;int main()
{int i, j, sm = 0,l = 0;int  x, y;while(scanf("%d %d",&x,&y)!=EOF){l++;int k = pow(y*10+3,1.0/3);sm = 0;for(i = x; i<=k; i++){for(j = x; j<=k; j++){if((i*i*i+j*j*j)%10==3&&(i*i*i+j*j*j)<=y*10+3)sm++;}}printf("Case %d: %d\n",l, sm);}
}