數據結構與算法之枚舉（窮舉）法 C++實現
枚舉法的本質就是從所有候選答案中去搜索正確的解，使用該算法需要滿足兩個條件：
1、可以先確定候選答案的數量；
2、候選答案的范圍在求解之前必須是一個確定的集合。
枚舉是最簡單，最基礎，也是最沒效率的算法
枚舉法優點：
1、枚舉有超級無敵准確性，只要時間足夠，正確的枚舉得出的結論是絕對正確的。
2、枚舉擁有天下第一全面性，因為它是對所有方案的全面搜索，所以，它能夠得出所有的解。
程序優化：
對於枚舉算法，加強約束條件，縮小枚舉的范圍，是程序優化的主要考慮方向。
百錢買百雞：
代碼：
#include
const int COCKPR = 3;
const int HENPR = 5;
const double CHICKPR = 1/3;
void buyChicken(int money, int chooks);
int main()
{
int money = 100;
int chooks = 100;
buyChicken(money, chooks);
return 0;
}
void buyChicken(int money, int chooks)
{
using namespace std;
int MaxCock = money/COCKPR;
int MaxHen = money/HENPR;
int MaxChick = chooks;
int cock,hen,chick;
int count = 0;
for(cock=0; cock<= MaxCock; cock++)
{
for(hen=0; hen<=MaxHen; hen++)
{
for(chick=0; chick<=MaxChick; chick++)
{
if (0 == chick%3 && cock + hen + chick == chooks && COCKPR*cock + HENPR*hen + CHICKPR*chick == money)
cout << "公雞: " << cock << " 母雞: " << hen << " 小雞: " << chick << " 第 " << count << "有結果" <
運行結果：
只對小雞的數量添加一個最小的起始值
函數裡面加 int MinChick = chooks-MaxCock-MaxHen;
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代碼如下：
#include
const int COCKPR = 3;
const int HENPR = 5;
const double CHICKPR = 1/3;
void buyChicken(int money, int chooks);
int main()
{
int money = 100;
int chooks = 100;
buyChicken(money, chooks);
return 0;
}
void buyChicken(int money, int chooks)
{
using namespace std;
int MaxCock = money/COCKPR;
int MaxHen = money/HENPR;
int MaxChick = chooks;
int MinChick = chooks-MaxCock-MaxHen;
int cock,hen,chick;
int count = 0;
for(cock=0; cock<= MaxCock; cock++)
{
for(hen=0; hen<=MaxHen; hen++)
{
for(chick=MinChick; chick<=MaxChick; chick++)
{
if (0 == chick%3 && cock + hen + chick == chooks && COCKPR*cock + HENPR*hen + CHICKPR*chick == money)
cout << "公雞: " << cock << " 母雞: " << hen << " 小雞: " << chick << " 第 " << count << "有結果" <
填數字游戲
代碼：
#include
int main()
{
using namespace std;
int t1,t2,t3,t4,t5;
for (t1=1; t1<=9; t1++)
{
for(t2=0; t2<=9; t2++)
{
for(t3=0; t3<=9; t3++)
{
for(t4=0; t4<=9; t4++)
{
for(t5=0; t5<=9; t5++)
{
if(t5*100000 + t5*10000 + t5*1000 + t5*100 + t5*10 + t5 == t5*t1 + t4*t1*10 + t3*t1*100 + t2*t1*1000 + t1 *t1*10000 )
{
cout << " 算的數值為: " << t1 << " 法的數值為: " << t2 << " 描的數值為: " <
運行結果：
