题目? ? ? ?? 实现一个经典"猜数字"游戏。?给定答案序列和用户猜的序列,统计有多少数字位置正确(A),有多少数字在两个序列都出现过但位置不对(B)。
输入 输入包含多组数据。?每组输入第一行为序列长度n,第二行是答案序列,接下来是若干猜测序列。?猜测序列全0时该组数据结束。?n=0时输入结束。
样例输入 4 1?3?5?5 1?1?2?3 4?3?3?5 6?5?5?1 6?1?3?5 1?3?5?5 0?0?0?0 10 1?2?2?2?4?5?6?6?6?9 1?2?3?4?5?6?7?8?9?1 1?1?2?2?3?3?4?4?5?5 1?2?1?3?1?5?1?6?1?9 1?2?2?5?5?5?6?6?6?7 0?0?0?0?0?0?0?0?0?0 0 样例输出 Game?1: (1,1) (2,0) (1,2) (1,2) (4,0) Game?2: (2,4) (3,2) (5,0) (7,0)
#include<stdio.h>
#define maxn 1010
int main() {
int n, a[maxn], b[maxn];
int kase = 0;
while (scanf("%d", &n) == 1 && n) {
printf("Game %d:\n", ++kase);
for (int i = 0; i < n; i++) scanf("%d", &a[i]);
for (;;) {
int A = 0, B = 0;
for (int i = 0; i < n; i++) {
scanf("%d", &b[i]);
if (a[i] == b[i]) A++;
}
if (b[0] = 0) break;
for (int d = 1; d <= 9; d++) {
int c1 = 0, c2 = 0;
for (int i = 0; i < n; i++) {
if (a[i] == d) c1++;
if (b[i] == d) c2++;
}
if (c1 < c2) B += c1; else B += c2;
}
printf(" (%d,%d)", A, B - A);
}
}
return 0;
}
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