猜数字游戏的提示（50页）

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int n,kase = 0;
 6     while (cin >> n,n != 0)
 7     {
 8         int a[500] = { 0 },b[100],mark =0,c1=0,c2=0;
 9         for (int i = 0; i < n; i++)
10             cin >> a[i];
11         while (true)
12         {
13             mark = 0;
14             for (int i = 0; i < n; i++)
15             {
16                 cin >> b[i];
17                 if (a[i] == b[i])
18                     mark++;
19             }
20             if (b[0] == 0) break;
21             int Sum = 0;
22             for (int i = 1; i <=9; i++)
23             {
24                 c1 = 0, c2 = 0;
25                 for (int j = 0; j < n; j++)
26                 {
27                     if (a[j] == i)    c1++;
28                     if (b[j] == i)    c2++;
29                 }
30                 if (c1 < c2) Sum += c1; else Sum += c2;
31             }
32             cout << "Game " << kase++ << endl;
33             cout << "(" << mark << "," << Sum - mark << ")" << endl;
34         }
35     }
36     return 0;
37 }

比较巧妙的是处理数字都存在但位置不对的方式，采取的方式是外层循环遍历1至9，用一个内层循环计两数组的数字的个数