HOJ 2201 Sum of Cubes_sum of cubes acm-优快云博客

本文链接：https://blog.youkuaiyun.com/cwallow/article/details/8542559

探索每一个整数是否能被分解为两个立方数之和，本文详细解析算法并提供实例解答。

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Time limit : 10 sec

Memory limit : 64 M

According to Goldbach’s conjecture, every evennumber can be expressed as a sum of two oddprimes. Which numbers can be expressed as thesum of two cubes?

For each test case, a line will contain a positive integer n which will be at most one million.For each test case, output two integers x and y such that x³ + y³ = n. If there are manysuch pairs of numbers, choose x to be as small as possible. If there are no such pairs, output“impossible”.

The last line will contain the integer 0, which should not be processed.

Sample Input

Sample Output

0 1
1 1
impossible
0 100

Solution:

#include <stdio.h>
#include <math.h>

int main()
{
        int x, y, n;
        while(1 == scanf("%d", &n))
        {
                if(0 == n)
                        return 0;
                x = -600;
                y = 600;
                for(;x <= y;)
                {
                        if ((int)(pow(x, 3) + pow(y, 3)) > n)
                        {
                                --y;
                        }
                        else if ((int)(pow(x, 3) + pow(y, 3)) < n)
                        {
                                ++x;
                        }
                        else
                        {
                                printf("%d %d\n", x, y);
                                break;
                        }
                }
                if (x > y)
                {
                        printf("impossible\n");
                }
        }
        return 0;
}

注意：x和y并没有指明是正数