猜数字游戏的提示（50页）

#include<iostream>
using namespace std;
int main()
{
    int n,kase = 0;
    while (cin >> n,n != 0)
    {
        int a[500] = { 0 },b[100],mark =0,c1=0,c2=0;
        for (int i = 0; i < n; i++)
            cin >> a[i];
        while (true)
        {
            mark = 0;
            for (int i = 0; i < n; i++)
            {
                cin >> b[i];
                if (a[i] == b[i])
                    mark++;
            }
            if (b[0] == 0) break;
            int Sum = 0;
            for (int i = 1; i <=9; i++)
            {
                c1 = 0, c2 = 0;
                for (int j = 0; j < n; j++)
                {
                    if (a[j] == i)    c1++;
                    if (b[j] == i)    c2++;
                }
                if (c1 < c2) Sum += c1; else Sum += c2;
            }
            cout << "Game " << kase++ << endl;
            cout << "(" << mark << "," << Sum - mark << ")" << endl;
        }
    }
    return 0;
}

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