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

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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
",k++ ,calc()) ;
    }
    return 0 ;
}