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HDU1334_Perfect Cubes【水题】

2024-08-03 16:36:10   224人阅读

Perfect Cubes


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2192    Accepted Submission(s): 957

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.

Source

Mid-Central USA 1995


题目大意：输出满足等式a^3 = b^3 + c^3 + d^3所有情况的a、b、c、d，

其中a<=200，a、b、c、d按非递减序列排列。

思路：直接遍历

#include<stdio.h>

int main()
{

    for(int a = 6; a <= 200; a++)
    {
        for(int b = 2; b < a; b++)
        {
            for(int c = b; c < a; c++)
            {
                for(int d = c; d < a; d++)
                    if(a*a*a == b*b*b + c*c*c + d*d*d)
                        printf("Cube = %d, Triple = (%d,%d,%d)\n",a,b,c,d);
            }
        }
    }

    return 0;
}


HDU1334_Perfect Cubes【水题】


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