POJ1543总结

Perfect Cubes

题目来源

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)

我的代码

先暴力a，在从小到大依次暴力b、c、d。这一题只有四个运算数以至于可以直接写四个循环。

#include<stdio.h>

int n,a,triple[4];
int cube[101];

void dfs(int step){
	if (step==3) {
		if (cube[a]==cube[triple[1]]+cube[triple[2]]+cube[triple[3]]){
			printf("Cube = %d, Triple = (%d,%d,%d)\n",a,triple[1],triple[2],triple[3]);
		}
	}
	else{
		for (int i=triple[step];i<a;i++){
			triple[step+1]=i;
			dfs(step+1);
		}
	}
}

int main()
{
	scanf("%d",&n);
	triple[0]=2;
	for (int i=0;i<=n;i++){
		cube[i]=i*i*i;
	}
	for (int i=3;i<=n;i++){
		a=i;
		dfs(0);
	}
	return 0;
}