The Ninth Hunan Collegiate Programming Contest (2013) Problem J

Problem J

Joking with Fermat's Last Theorem

Fermat's Last Theorem: no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two.

From the theorem, we know that a3 + b3 = c3 has no positive integer solution.

However, we can make a joke: find solutions of a3 + b3 = c3. For example 43 + 93 = 793, so a=4, b=9, c=79 is a solution.

Given two integers x and y, find the number of solutions where x<=a,b,c<=y.

Input

There will be at most 10 test cases. Each test case contains a single line: x, y (1<=x<=y<=108).

Output

For each test case, print the number of solutions.

Sample Input

1 10
1 20
123 456789

Output for the Sample Input

Case 1: 0
Case 2: 2
Case 3: 16

The Ninth Hunan Collegiate Programming Contest (2013) Problemsetter: Rujia Liu Special Thanks: Md. Mahbubul Hasan, Feng Chen

这道题有一个突破口，就是a, b <=1000 ,这样子算法就变成了 O（1000*1000）。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
LL N_3[1001] ;
int  x , y ;
void init(){
for(int i=1;i<=1000;i++)
N_3[i]=i*i*i  ;
// cout<<N_3[1000]<<endl ;
}
int calc(){
int a_up ,b_up ,c ,sum ,ans=0 ;
a_up=Min(y,1000) ;
b_up=Min(y,1000) ;
for(int a=x;a<=a_up;a++)
for(int b=x;b<=b_up;b++){
sum=N_3[a]+N_3[b] ;
if(sum%10==3){
int c=sum/10 ;
if(x<=c&&c<=y)
ans++ ;
}
}
return ans ;
}
int main(){
init() ;
int k=1 ;
while(scanf("%d%d",&x,&y)!=EOF){
printf("Case %d: %d\n",k++ ,calc()) ;
}
return 0 ;
}