CSU 1337 搞笑版费马大定理（2013湖南省程序设计竞赛J题）

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdlib>
 6 using namespace std;
 7 typedef long long INT;
 8 INT th[1005];
 9 void dabiao()
10 {
11     for(INT i = 1;i <= 1000;++i)
12     th[i] = i * i * i;
13 }
14 int main()
15 {
16     dabiao();
17     INT x,y;
18     int kase = 1;
19     while(scanf("%lld%lld",&x,&y)!=EOF)
20     {
21         INT ans = 0;
22         INT a = min(x,(INT)1000);    //不存在大于1000的 
23         INT b = min(y,(INT)1000);
24         for(INT i = a;i <= b;++i)
25         for(INT j = a;j <= b;++j)
26         {
27             INT s = th[i] + th[j];
28             if(s % 10 == 3 && s / 10 <= y)
29             ans++;
30         }
31         printf("Case %d: %lld\n",kase++,ans);
32     }
33     return 0;
34 }