HDOJ 1334 Perfect Cubes(暴力)

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## HDOJ 1334 Perfect Cubes(暴力)

Problem 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 <= 200.

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.

The first part of the output is shown here:

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)

Note: The programmer will need to be concerned with an efficient implementation. The official time limit for this problem is 2 minutes, and it is indeed possible to write a solution to this problem which executes in under 2 minutes on a 33 MHz 80386 machine. Due to the distributed nature of the contest in this region, judges have been instructed to make the official time limit at their site the greater of 2 minutes or twice the time taken by the judge’s solution on the machine being used to judge this problem.

（<2a<=b<=c<200）

``````public class Main{
public static void main(String[] args) {
for(int m=6;m<=200;m++){

int mt = m*m*m;
int at;
int bt;
int ct;
for(int a=2;a<m;a++){
at=a*a*a;

for(int b=a;b<m;b++){
bt = b*b*b;
//适当的防范一下，提高效率
if(at+bt>mt){
break;
}

for(int c=b;c<m;c++){
ct=c*c*c;

//适当的防范一下，提高效率
if(at+bt+ct>mt){
break;
}

if(mt==(at+bt+ct)){
System.out.println("Cube = "+m+", Triple = ("+a+","+b+","+c+")");
}

}
}
}
}

}

}
``````

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