# The theory of Fei Ma, the funny edition of Hunan's ninth College Students' Computer Programming Competition

Source: Internet
Author: User

Funny free horse theorem time limit: 1 sec memory limit: 128 MB
Submit: 190 solved: 93
[Submit] [Status] [web board] Description

Ferma's theorem: WHEN n> 2, there is no positive integer solution for the Indefinite Equation an + BN = cn. For example, A3 + B3 = C3 has no positive integer solution. For an active atmosphere, let's make a funny version: change the equation to A3 + B3 = C3, so there is a solution, such as a = 4, B = 9, C = 79 at 43 + 93 = 793.

Enter two integers x and y to calculate the number of integer solutions that meet the conditions of x <= A, B, C <= y.

Input

The input can contain up to 10 groups of data. Each group of data contains two integers x and y (1 <= X, Y <= 108 ).

Output

The number of outputs for each group of data.

Sample Input
`1 101 20123 456789`
Sample output
`Case 1: 0Case 2: 2Case 3: 16`
Hintsource

Hunan ninth College Computer Program Design Competition

Enumeration..., mainly to block out the range, otherwise timeout

The AC code is as follows:

`#include<iostream>#include<cmath>#include<cstring>#include<stdio.h>using namespace std;int main(){    int x,y;    int a,b,c;    int i,j;    int cas=1;    int l,ans,sum,bj;    while(~scanf("%d%d",&x,&y))    {        sum=0;bj=0;        for(i=x;i*i*i<=y*10;i++)        {            for(j=i;j*j*j<=y*10;j++)            {                ans=i*i*i+j*j*j;                if(ans%10==3&&ans/10>=x&&ans/10<=y)                {                    sum+=2;                    //cout<<i<<" "<<j<<endl;                    if(i==j)                        sum--;                }                if(2*i*i*i/10>y)                {break;bj=1;}            }            if(bj)                break;        }        cout<<"Case "<<cas++<<":"<<" ";        cout<<sum<<endl;    }     return 0;}`

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