数学--数论--HDU-2698 Maximum Multiple（规律）

Given an integer nn, Chiaki would like to find three positive integers xx, yy and zzsuch that: n=x+y+zn=x+y+z, x∣nx∣n, y∣ny∣n, z∣nz∣n and xyzxyz is maximum.

Input

There are multiple test cases. The first line of input contains an integer TT (1≤T≤1061≤T≤106), indicating the number of test cases. For each test case:

The first line contains an integer nn (1≤n≤1061≤n≤106).

Output

For each test case, output an integer denoting the maximum xyzxyz. If there no such integers, output −1−1 instead.

Sample Input

Sample Output

-1

-1

1

只有因子中有4或者有3才能被拆成 X+Y+Z=N,然后打了表验证。

最后wa了好几次，是因为int和int计算之后还是int就算赋值给long long .

打表代码

#include <bits/stdc++.h>

using namespace std;

int main()

{

	int T;

	scanf("%d", &T);

	while (T--)

	{

		int n;

		for (int n = 1; n <= 100; n++)

		{

			int maxt = -1;

			int a, b, c;

			for (int x = 1; x <= n; x++)

			{

				for (int y = 1; y <= n - x; y++)

				{

					int z = n - x - y;

					if (z && n % x == 0 && n % y == 0 && n % z == 0)

					{

						if (maxt < x * y * z)

						{

							a = x;

							b = y;

							c = z;

						}

						maxt = max(maxt, x * y * z);

					}

				}

			}

			printf("%d:%5d %d %d %d\n", n, maxt, a, b, c);

			if (n % 12 == 0)

				printf("\n");

		}

	}

	return 0;

}

#include <bits/stdc++.h>

using namespace std;

int main()

{

	int T;

	long long  n,x,y,z;

	long long sum;

	scanf("%d", &T);

	while (T--)

	{

		scanf("%lld", &n);

		if ((n % 3) == 0)

		{

			x = y = z = n / 3;

			sum = x * y * z;

			if (x + y + z == n)

				printf("%lld\n", sum);

			else

				puts("-1");

		}

		else if ((n % 4) == 0)

		{

			x = y = n / 4, z = n / 2;

			sum = x * y * z;

			if (x + y + z == n)

				printf("%lld\n", sum);

			else

				puts("-1");

		}

		else

			puts("-1");

	}

}

巴特西

数学--数论--HDU-2698 Maximum Multiple（规律）

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