HDU1334 ZOJ1331 UVA386 UVALive5303 Perfect Cubes【暴力】

Perfect Cubes

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3803    Accepted Submission(s): 1781

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

Regionals 1995 >> North America - Mid-Central USA

问题链接：HDU1334 ZOJ1331 UVA386 UVALive5303 Perfect Cubes

问题简述：（略）

问题分析：

　　这个问题适合暴力枚举，更加巧妙的方法也难以想出来。

　　即便是枚举，也需要考虑能省则省。

　　这个题是原题，一个变种参见参考链接。只是这个原题没有输入，a、b、c和d值最大为200。

程序说明：

　　使用立方数组，可以减少重复的立方计算。

　　使用变量sum，可以减少重复的求和变量。

　　当和sum大于a^3时，就不需要再试探下去了。b^3+c^3大于a^3时，同样不用再试探，结束循环。

　　b、c和d的值根据题意，1

　　所以2<=b<=c<=d，得24<=b^3+c^3+d^3，取3<=a开始计算。

题记：（略）

参考链接：POJ1543 Perfect Cubes【暴力】

AC的C语言程序如下：

/* HDU1334 ZOJ1331 UVA386 UVALive5303 Perfect Cubes */

#include 

#define N 200
long cube[N + 1];

void setcube(int n)
{
    int i;

    for(i=0; i<=n; i++)
        cube[i] = i * i * i;
}

int main(void)
{
    setcube(N);

    long a, b, c, d, sum;

    for(a=3; a<=N; a++)
        for(b=2; b<=N; b++)
            for(c=b; c<=N; c++) {
                if(cube[b] + cube[c] > cube[a])
                    break;
                for(d=c; d<=N; d++) {
                    sum = cube[b] + cube[c] + cube[d];
                    if(cube[a] == sum)
                        printf("Cube = %ld, Triple = (%ld,%ld,%ld)\n", a, b, c, d);
                    else if(sum > cube[a])
                        break;
                }
            }

    return 0;
}