計算24点の簡単なプログラム

最近の宿題には24時の問題があり、可能な組み合わせを見つけるために簡単なプログラムを書きました.
1.プログラムの実行
2.整数を4つ入力します.たとえば、3 3 3 7 8
3.可能なすべての組合せを表示

#include "assert.h"
#include 

double operate(double num1, double num2, int op)
{
	assert(op >= 0 && op < 4);

	if(op == 0){
		return num1 + num2;
	}
	else if(op == 1){
		return num1 - num2;
	}
	else if(op == 2){
		return num1 * num2;
	}
	else{
		return num1/num2;
	}
}

int calculate(int num1, int num2, int num3, int num4)
{
	char operators[] = "+-*/";

	for(int i = 0; i < 4; i ++)
	{
		for(int j = 0; j < 4; j ++)
		{
			for (int k = 0; k < 4; k ++)
			{
				double ret = operate(num1, num2, i);
				ret = operate(ret, num3, j);
				ret = operate(ret, num4, k);
				if(abs(ret - 24) < 0.001){
					printf("((%d %c %d) %c %d) %c %d = %f
", num1, operators[i], 
						                                  num2, operators[j], 
										  num3, operators[k], 
										  num4, ret);
				}

				ret = operate(num1, num2, i);
				double ret2 = operate(num3, num4, k);
				ret = operate(ret, ret2, j);
				if(abs(ret - 24) < 0.001){
					printf("(%d %c %d) %c (%d %c %d) = %f
", num1, operators[i], 
						                                  num2, operators[j], 
										  num3, operators[k], 
										  num4, ret);
				}

				ret = operate(num2, num3, j);
				ret = operate(num1, ret, i);
				ret = operate(ret, num4, k);
				if(abs(ret - 24) < 0.001){
					printf("(%d %c (%d %c %d)) %c %d = %f
", num1, operators[i], 
						                                  num2, operators[j], 
										  num3, operators[k], 
										  num4, ret);
				}

				ret = operate(num2, num3, j);
				ret = operate(ret, num4, k);
				ret = operate(num1, ret, i);
				if(abs(ret - 24) < 0.001){
					printf("%d %c ((%d %c %d) %c %d) = %f
", num1, operators[i],
						                                  num2, operators[j], 
										  num3, operators[k], 
										  num4, ret);
				}

				ret = operate(num3, num4, k);
				ret = operate(num2, ret, j);
				ret = operate(num1, ret, i);
				if(abs(ret - 24) < 0.001){
					printf("%d %c (%d %c (%d %c %d)) = %f
", num1, operators[i], 
						                                  num2, operators[j], 
										  num3, operators[k], 
										  num4, ret);
				}
			}
		}
	}
	return 0;
}

int main(int argc, char* argv[])
{

	int nums[4] = {0, 0, 0, 0};
	std::cin >> nums[0] >> nums[1] >> nums[2] >> nums[3];

	for (int i = 0; i < sizeof(nums)/sizeof(nums[0]); i ++)
	{
		int num1 = nums[i];
		int ret = num1;
				
		for(int j = 0; j < sizeof(nums)/sizeof(nums[0]); j ++)
		{
			if(j == i)
				continue;

			int num2 = nums[j];
			
			for(int k = 0; k < sizeof(nums)/sizeof(nums[0]); k++)
			{
				if( k == i || k == j)
					continue;

				int num3 = nums[k];

				for(int l = 0; l < sizeof(nums)/sizeof(nums[0]); l ++)
				{
					if(l == i || l == j || l == k)
						continue;

					int num4 = nums[l];
					calculate(num1, num2, num3, num4);
				}
			}
		}
	}

	return 0;
}