假設rand5能隨機生成1~5的5個數(均等概率)，利用rand5生成rand7() 1~7(均等概率)

1. 利用rand5求出rand2(),當rand5生成的數大於2時，一直迴圈，直到它生成的數為1或者2，且生成1和2的概率均為1/2，因為生成1和生成2的概率是相等的。

2. 利用rand2生成0和1（減1即可），連續使用三次，共有8種可能（二進位制表示）：000 001 010 011 100 101 110 111，當生成的數為000時，重新計算，這樣就可以得到1~7範圍內的數，且概率是均等的。

#include <iostream>
#include <math.h>

using namespace std;

int rand5()
{
	return rand()%5+1;
}

int rand2()
{
	int temp;
	do 
	{
		temp = rand5();
	} while (temp > 2);

	return temp;
}

int rand7()
{
	int temp;
	do 
	{
		temp = (rand2()-1)<<2;
		temp += (rand2()-1)<<1;
		temp += rand2()-1;
	} while (temp == 0);

	return temp;
}

int main(int, char **)
{
	int num;
	cout << "輸入總生成數:";
	cin >> num;

	int c1, c2, c3, c4, c5, c6, c7;
	c1 = c2 = c3 = c4 = c5 = c6 = c7 = 0;
	for (int i = 0; i < num; i++)
	{
		int temp = rand7();
		switch (temp)
		{
		case 1:
			c1++;
			break;
		case 2:
			c2++;
			break;
		case 3:
			c3++;
			break;
		case 4:
			c4++;
			break;
		case 5:
			c5++;
			break;
		case 6:
			c6++;
			break;
		case 7:
			c7++;
		}

		//cout << temp << "\t";
	}

	cout << endl;
	cout << "生成1的個數佔總生成數的：" << (double)c1/num*100 << "%" << endl;
	cout << "生成2的個數佔總生成數的：" << (double)c2/num*100 << "%" << endl;
	cout << "生成3的個數佔總生成數的：" << (double)c3/num*100 << "%" << endl;
	cout << "生成4的個數佔總生成數的：" << (double)c4/num*100 << "%" << endl;
	cout << "生成5的個數佔總生成數的：" << (double)c5/num*100 << "%" << endl;
	cout << "生成6的個數佔總生成數的：" << (double)c6/num*100 << "%" << endl;
	cout << "生成7的個數佔總生成數的：" << (double)c7/num*100 << "%" << endl;

	return 0;
}