//例题3-4 猜数字游戏的提示
//收获:处理思路:在猜测和答案中都出现某数字,且在猜测和答案位置不同的次数 = 都出现的次数 - 数字位置正确的次数; 都出现的次数 = min(答案中出现次数,猜测串中出现次数)
#include <iostream>
using namespace std;
int main()
{
int n, *a, *b;
int sum1, sum2, c1, c2, kase = 0;
while (cin >> n)
{
if (!n) break;
a = new int [n];
b = new int [n];
cout << "Game " << ++kase << ":" << endl;
for (int i = 0; i < n; i++) cin >> a[i];
while(1)
{
sum1 = sum2 = 0;
for (int i = 0; i < n; i++)
{
cin >> b[i];
if (a[i] == b[i]) sum1++;
}
if (b[0] == 0) break;
for (int j = 1; j <= 9; j++)
{
c1 = c2 = 0;
for (int i = 0; i < n; i++)
{
if (a[i] == j) c1++;
if (b[i] == j) c2++;
}
sum2 += min(c1, c2);
}
cout << " (" << sum1 << "," << sum2 - sum1 << ")" << endl;
}
}
delete[]a;
delete[]b;
return 0;
}
UVA - 340 Master-Mind Hints
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2024-02-24 23:16:16
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