标签：des style blog io ar os sp java for
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】
标签：des style blog io ar os sp java for
原文地址：http://blog.csdn.net/lianai911/article/details/41358227
