ZJU 1331 解题报告

Perfect CubesTime limit: 10 Seconds Memory limit: 32768K
Total Submit: 2409 Accepted Submit: 1045 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.

Problem Source: Mid-Central USA 1995

题目地址:http://acm.zju.edu.cn/show_problem.php?pid=1331

初看题目，首先想到的就是暴力解决，四个for循环，代码如下：


#include <stdio.h>




int main(int argc, char *argv)




...{


int a, b, c, d;




for (a = 2; a <= 200; a ++)


for (b = 2; b < a; b ++)


for (c = b; c < a; c ++)


for (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) ", a, b, c, d);


break;


}




return 0;


}

但这样做，显然效率太低，每个数的立方都要重复计算很多次，可以定义一个数组，以空间换时间，代码如下：


#include <stdio.h>




int main(int argc, char *argv)




...{


int a, b, c, d, i, cube[201];




for (i = 0; i < 201; i ++) cube[i] = i * i * i;


for (a = 2; a <= 200; a ++)


for (b = 2; b < a; b ++)


for (c = b; c < a; c ++)


for (d = c; d < a; d ++)




if (cube[a] == cube[b] + cube[c] + cube[d]) ...{


printf("Cube = %d, Triple = (%d,%d,%d) ", a, b, c, d);


break;


}




return 0;


}

再优化，程序中的加法有很多是重复的，一些循环控制的初值也可改变以减少循环次数。如下：


#include <stdio.h>




int main(int argc, char *argv)




...{


int a, b, c, d, i, x, y, cube[201];




for (i = 1; i < 201; i ++) cube[i] = i * i * i;


for (a = 6; a <= 200; a ++)




for (b = 2; b < a; b ++) ...{


x = cube[a] - cube[b];




for (c = b + 1; c < a; c ++) ...{


y = x - cube[c];


for (d = c + 1; d < a; d ++)




if (y == cube[d]) ...{


printf("Cube = %d, Triple = (%d,%d,%d) ", a, b, c, d);


break;


}


}


}




return 0;


}

最后一个for循环的if判断语句为顺序查找，对于一个已经有序的数组来说效率是比较低的，可以改为二分查找。如下：


#include <stdio.h>




int BinarySearch(int s[], int high, int low, int key);




int main(int argc, char *argv)




...{


int a, b, c, d, i, cube[201];




for (i = 1; i < 201; i ++) cube[i] = i * i * i;


for (a = 6; a <= 200; a ++)




for (b = 2; b < a; b ++) ...{


i = cube[a] - cube[b];




for (c = b + 1; c < a; c ++) ...{


d = BinarySearch(cube, a - 1, c + 1, i - cube[c]);


if (d) printf("Cube = %d, Triple = (%d,%d,%d) ", a, b, c, d);


}


}




return 0;


}




int BinarySearch(int s[], int high, int low, int key)




...{


int mid;






while (low <= high) ...{


mid = (high + low) / 2;


if (s[mid] == key) return mid;


else if (s[mid] < key) low = mid + 1;


else high = mid - 1;


}




return 0;


}

至此，已想不出可以如何再优化，提交：

[align=center]Judge Status[/align]	[align=center]Problem ID[/align]	[align=center]Language[/align]	[align=center]Run time[/align]	[align=center]Run memory[/align]
[align=center]Accepted[/align]	[align=center]1331[/align]	[align=center]C[/align]	[align=center]00:00.10[/align]	[align=center]388K[/align]