Perfect Cubes
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 14901 | Accepted: 7804 |
Description
For hundreds of years Fermat's Last Theorem, which stated simply that for n > 2 there exist no integers a, b, c > 1 such that a^n = b^n + c^n, has remained elusively unproven. (A recent proof is believed to be correct, though it is still undergoing scrutiny.) It is possible, however, to find integers greater than 1 that satisfy the "perfect cube" equation a^3 = b^3 + c^3 + d^3 (e.g. a quick calculation will show that the equation 12^3 = 6^3 + 8^3 + 10^3 is indeed true). This problem requires that you write a program to find all sets of numbers {a,b,c,d} which satisfy this equation for a <= N.
Input
One integer N (N <= 100).
Output
The output should be listed as shown below, one perfect cube per line, in non-decreasing order of a (i.e. the lines should be sorted by their a values). The values of b, c, and d should also be listed in non-decreasing order on the line itself. There do exist several values of a which can be produced from multiple distinct sets of b, c, and d triples. In these cases, the triples with the smaller b values should be listed first.
Sample Input
24
Sample Output
Cube = 6, Triple = (3,4,5)
Cube = 12, Triple = (6,8,10)
Cube = 18, Triple = (2,12,16)
Cube = 18, Triple = (9,12,15)
Cube = 19, Triple = (3,10,18)
Cube = 20, Triple = (7,14,17)
Cube = 24, Triple = (12,16,20) 题意：找出2到n中，所有满足a^3=b^3+c^3+d^3的a,b,c,d的数
#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 2002000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int main()
{
int n,m,j,i,t,k,l;
int vis[1000];
int a[150];
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
a[i]=pow(i,3);
for(i=6;i<=n;i++)
{
memset(vis,0,sizeof(vis));
for(j=2;j<i;j++)
{
if(a[i]<a[j]+a[j+1]+a[j+2])
break;
for(k=j;k<i;k++)
{
if(a[i]<a[j]+a[k]+a[k+1]) break;
for(l=k;l<i;l++)
{
if(a[i]==a[j]+a[k]+a[l])
{
printf("Cube = %d, Triple = (%d,%d,%d)\n",i,j,k,l);
}
}
}
}
}
}
return 0;
}