暴力求解法

一.简单枚举

第一题：除法
#include <cstdio>
#include <iostream>

using namespace std;

void dd(int n)
{
	
	for (int i = 98765 / 79; i <= 98765 / 2; i++)//i是被除数
	{
		int y[12] = { 0 };
		if (i < 10000)
		{
			y[0] = 1;

		}
		int k = i;//对当前数进行分解
		int p;
		bool isok = true;
		do {
			int temp = k % 10;
			if (y[temp] == 1)
			{
				isok = false; break;
			}
			else {
				y[temp] = 1;
			}
			k /= 10;
		} while (k);
		if (isok == false)
		{
			continue;
		}
		else
		{
			p = n * i;
			int pp = p;
			do {
				int temp = pp % 10;
				if (1 == y[temp]) {
					isok = false; break;
				}
				else {
					y[temp] = 1;
				}
				pp /= 10;
			} while (pp);
		}
		if (isok)
		{
			if (i < 10000)
			{
				printf("%d/0%d=%d\n", p, i, n);
			}
			else {
				printf("%d/%d=%d\n",p, i, n);
			}
		}
	}
}
int main()
{
	int n;
	cin >> n;
	//scanf_s("%d", &n);
	dd(n);
	
	return 0;
}


第二题：最大乘积
int main() {
	int n;
	cin >> n;
	int aa[20]; int i;
	for (i = 0; i < n; i++)
	{
		cin >> aa[i];
	}
	long long max = aa[0] * aa[1];
	for (int begin = 0; begin <=i- 1; begin++)
	{
		for (int end = begin + 1; end <=i; end++)
		{
			long long p = 1;
			for (int j = begin; j <end; j++)
			{
				p *= aa[j];
			}
			if (p > max)
			{
				max = p;
			}
		}
	}
	if (max < 0)
	{
		cout << "0" << endl;
	}
	else {
		cout << max << endl;
	}
	return 0;
}


第三题：分数拆分
int main()
{
	int k;
	cin >> k;
	int ans = 0;
	int y, x;
	for (y = k + 1; y <= 2 * k; y++)
	{
		x = k*y / (y - k);
		if (k * y %(y - k)==0)//保证是个整数
		{
			++ans; 
			cout<<ans<<endl;
			printf("1/%d=1/%d+1/%d\n", k, x, y);
		}
	}
	return 0;
}

暴力求解法

原文：https://www.cnblogs.com/lwyy1223-/p/13558546.html