POJ 1543 Perfect Cubes（我的水题之路——四重暴力水）

Perfect Cubes

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10707 Accepted: 5739

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 <= N.

Input

One integer N (N <= 100).

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.

Sample Input

24

Sample Output

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)

Source

Mid-Central USA 1995

要求寻找N（N<=100）以下的，a^3=b^3+c^3+d^3的所有组合。要求按照以a为升序输出，同时，(b,c,d)中必须有b<=c<=d

四重暴力，不解释。

注意点：

1）如果四重暴力时候，上限均取为100，那么在查找数组的时候，循环条件如果是A<=N，必须考虑到N=100的情况

代码（1AC）：

#include <cstdio>#include <cstdlib>int answer[100][4];int top = 0;void initsolve(){    int i, j;    int a, b, c, d;    for (a = 2; a <= 100; a++){        for (b = 2; b <= 100; b++){            for (c = b; c <= 100; c++){                for (d = c; d <= 100; d++){                    if (a * a * a == b * b * b + c * c * c + d * d * d){                        answer[top][0] = a;                        answer[top][1] = b;                        answer[top][2] = c;                        answer[top][3] = d;                        top++;                        //printf("a=%d, b=%d, c=%d, d=%d\n", a, b, c, d);                    }                }            }        }    }}int main(void){    int N;    int i;    initsolve();    while (scanf("%d", &N) != EOF){        for (i = 0; i < top && answer[i][0] <= N; i++){            printf("Cube = %d, Triple = (%d,%d,%d)\n", answer[i][0], answer[i][1], answer[i][2], answer[i][3]);        }    }    return 0;}